solve the quantity x plus three times
the quantity x minus five
equals negative nineteen let's begin by
writing this in standard form equal to
zero
to do that we need to apply the
distributive property
in this case we have x squared
minus 5x plus 3x
minus 15 equals
negative 19. on the left side here we
have
like terms combining those we'll be left
with
x squared minus 2x
minus 15 equals negative 19.
and then finally adding 19 to both sides
we have
standard four x squared
minus two x plus four
equals zero
once we have standard form we can then
identify a b
and c for use in the quadratic formula
so in this case a is 1
b is negative 2 and c the constant term
here
is 4. we're going to plug these values
into the quadratic formula which is
x equals negative b plus or minus
the square root of b squared minus 4
ac all divided by
2 times a now substituting in the values
we have
negative b or negative negative 2 in
this case
plus or minus the square root
of negative two squared
minus four times a ac so four times a
which is one
c is four
all divided by 2 times a a is 1.
now simplifying further we have negative
negative 2 which is positive 2
plus or minus the square root
negative two squared is four so four
minus sixteen
all divided by two
continuing we have 2 plus or minus
the square root 4 minus 16 is
negative 12 divided by 2.
now at this point we can write negative
12 as the square root
of negative 4 times 3
and the square root of negative 4 is
2 i so we have 2 plus or minus
2i times the square root of 3 divided by
2.
and then finally reducing
2 divided by 2 leaves us with a 1 plus
or minus
2i square root of 3 divided by 2
leaves us with the term i times the
square root of 3.
so these are the two solutions to the
original equation
notice they're both complex
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