This is the general form of a
quadratic equation.
a x square plus b x plus c equal to 0.
a cannot be 0 because if a is 
zero this part goes away
and it is no longer a quadratic equation. 
So let us solve  
this quadratic equation using the method
of completing the square.
We have learnt that we can apply
the method of completing the square
in two manners. We require 
this term to be a perfect square. So we
can make it
a perfect square either by diving 
the entire equation by a.
So if we divide it by a 
this becomes x square.
Or we can multiply the entire 
equation by a.
So if we multiply the entire 
equation by a
this becomes a square x square
which is a perfect square.
So let us solve this by both the methods.
First let us take this constant term
to the right hand side.
So taking this constant 
term to the right hand side
we get a x square plus b x equal 
to negative c.
Now let us divide both the sides 
of this equation by a.
So divide all the terms by a.
So we get x square 
Plus b by a into x
equal to negative c by a.
This is what we have.
Now this can be written 
as x whole square 
Now our aim is to make a perfect 
square on the left hand side.
So we have x whole square.
Now we write this as 2 into
we take this x into 
b by 2a. 
b by a into x is the same as 2 into x into b by 2a.
And now we take b by 2a whole square.
So this becomes
x square plus 2 into x 
into b by 2a plus b by 2a whole square.
Since we have added b by 2a whole 
square to the left hand side 
we add this to the right hand 
side as well.
So we write the right hand side 
as minus c by a plus
b by 2a whole square.
This is what we have.
Now the left hand side can 
be written as 
x plus b by 2a
whole square.
This is x plus b by 2a whole square.
And the right hand side is 
negative c by a
plus
b square by 4 a square.
So this is what we have.
x plus b by 2a whole square is equal to 
negative c by a plus b square by 4a square.
Now simplify the right hand side.
So we take the LCM. Taking the LMC
we get 4 a square
a into 4a is 4 a square.
So multiply negative c by 4a
We get -4ac
plus b square
So we have b square minus 4ac by 
4 a square on the right hand side.
So this is what we have.
Now a is a constant b is a constant
c is a constant. So the right hand side
is a constant.
The left hand side is in the 
form of something square.
So we can solve it by taking the 
square root of both the sides.
So we have this. Now
let us solve it by taking the 
square root of both the sides.
So taking the square root of both the sides
we get this
root over x plus b by 2a whole square
will be x plus b by 2a and we
consider the positive and negative
square root of the right hand side.
This becomes
plus minus root over 
b squared minus 4ac.
That is square root of numerator divided 
by square root of denominator.
Now
Root over 4 a square will be 2a.
So this is what we have on the RHS. 
x plus b by 2a
is equal to plus/minus root over 
b square minus 4ac by 2a.
Now
take this to the right hand side.
So we can write x is equal to
negative b by 2a
plus minus root over b square minus 4ac
by 2a.
So this is what we have.
Now we have to 2a in the denominator 
in both the terms.
So we take the Lcm and we write it
as 2a
negative b plus minus
root over b square minus 4ac.
So this is the value of x.
So this is the solution for x
for the general form of a quadratic
equation.
x is equal to minus b
plus minus root over 
b square minus 4ac
upon 2a. These are the two values of x.
One is negative b plus root over b square 
minus 4ac by 2a and the other
is
negative b minus root over b 
square minus 4ac by 2a
So these are the two values of x.
Now let us solve this general 
form of a quadratic equation
by multiplying the entire equation by a.
First let us take this constant 
term to the right hand side.
So we write it as a x square plus b x
equal do negative c.
Now we multiply both the sides by a.
So this becomes a square into x square.
b x into a is abx.
Negative c into a is –ac.
So this is what we have.
Now this is
ax whole square 
plus now we need to write it 
as 2 into this term
ax
into b by 2.
2 into ax into b by 2 is the same as abx.
So we have a x whole square here.
We require a b by 2 whole square.
So we add a b by 2 whole square.
Since we have added it to the 
LHS we need to add it to 
the RHS as well. So we add b by 2 whole 
square.
So this is what we have. We add b by 2 
whole square to both the sides.
Now the left hand side is ax
plus b by 2 whole square. 
So left hand side is ax plus 
b by 2 whole square.
And the right hand side is 
negative ac plus b by 2 whole square.
Now b square and 2 square is 4.
So negative ac plus b square by 4.
So we have we take the LCM and we write 
it as b square minus 4ac by 4.
So b square by 4 minus 
we multiply this by 4
-4ac by 4.
Now a is a constant b is a constant 
c is a constant. So this entire 
right hand side is constant.
On the left hand side we have something
square.
So we have obtained the square of 
something is equal to some constant.
Now we can solve 
this by taking the square root. 
So let us take the square 
root of both the sides.
On this side we need to add the positive
and negative sign.
So taking the square root 
we get ax plus b by 2.
on the left hand side and 
on the right hand side 
the entire thing goes under the root
and we have a positive and negative
sign.
Now 
This is equal to the square 
root of numerator
divided by the square 
root of denominator. 
So root over 4
is 2. So this is what we have here.
Now we can take this b by 2 
to the right hand side and subtract it.
So we have ax is equal to negative b by 2
plus minus root over b square 
minus 4ac by 2.
So we have this ax here.
We can write it as negative b plus
minus root over b square minus 4ac 
by 2.
Now we need to find the value of x.
Now 
we transpose this a to the denominator
of the right hand side.
So x is equal to minus b 
plus minus root over b square minus 4ac
by 2a. So this is
the solution for x. Note
that this is the solution for x 
for the general form of a quadratic
equation.
So now given any quadratic equation 
you can compare it to the general 
form of a quadratic equation 
and use this formula directly to
solve the quadratic equation.
This formula is a very important
formula in mathematics. It is known as
the quadratic formula
or the Sridaracharayas formula. 
Why quadratic formula?
Because it is used to solve a quadratic
equation
of the form a x square plus bx 
plus c equal to 0.
It is known as the 
Sridaracharayas formula
because Sridaracharaya was the Indian
mathematician
who first gave this formula. So you
should always remember this formula
as it is very important for solving
quadratic equations.
Now let us solve an example 
using this.
So this is a quadratic equation. We need
to compare
it to the general form of 
a quadratic equation 
which is this and then 
apply this formula.
So the coefficient of x square is 
2. So we have a
is equal to 2.
The coefficient of x is 5. So b
is equal to 5
and the constant term is -6 so c
is equal to -6.
So this is what we have. Now
we simply need to put these values 
in this general form.
So use this and put the 
values of the a b and c.
So we have x is equal to
minus b so minus
b here is 5
plus minus plus minus
root over
b square. b is 5 so we have 5 square.
Then we have a minus sign. Minus
4 a into c.
a is 2 and c
is -6.
So this is what we have. And in the
denominator
we have 2 into a. So we
write 2 into a is 2
2 into 2.
So this is what we have.
Now let us simplify this.
This is -5 plus minus
root over five square is 25.
This negative and this 
negative become positive.
4 into 2 is 8. 8 into 6
is 48
divided by 2 into 2 that is 4.
So this is what we have.
Or this is equal to -5 plus minus
root over 25 plus 48
is 73.
25 plus 48 is 73.
So 5 plus minus 73 by 4. 
So these are the two solutions of x.
How two ? Because this is
-5 plus root over 73 by 4
or -5 minus root over 73 by 4.
So we have got this solution using 
the Sridaracharya's formula
or the quadratic formula.
