Question

-3x^{2}-21x-54 what are the zeros? (Solutions)

Answer 1

The only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54

remember if we can do
xy=0 then assume x and y=0

so factor

0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is

if you have
ax^2+bx+c=0 then
x=$\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$

x^2+7x+18
a=1
b=7
c=18

x=$\frac{-7+/- \sqrt{7^{2}-4(1)(18)} }{2(1)}$
x=$\frac{-7+/- \sqrt{49-72} }{2}$
x=$\frac{-7+/- \sqrt{-23} }{2}$
i=√-1
x=$\frac{-7+/- i\sqrt{23} }{2}$



the zerose would be
x=$\frac{-7+ i\sqrt{23} }{2}$ or $\frac{-7- i\sqrt{23} }{2}$