How do you find the roots, real and imaginary, of $y=-2(x +3)^2-7x+5$ using the quadratic formula?

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Answer 1 · 2017-08-30T03:32:59

$x=−8.75780488547035$ , $x=−0.7421951145296504$

There are no imaginary solutions.

First, combine like terms and simplify the expression so it's in standard quadratic form:

$ax^2+bx+c$

The original expression:

$-2(x+3)^2-7x+5$

After FOILing $(x+3)$ :

$(x+3)(x+3)$

$=x^2+6x+9$

After multiplying that by $-2$ :

$-2x^2-12x-18$

After adding additional terms and simplifying:

$-2x^2-19x-13$

To find the roots (AKA zeros), just plug it into the quadratic formula:

$x=(-b\pm\sqrt{b^2-4ac})/(2a)$

$=(9\pm\sqrt{(-19)^2-4(-2)(-13)})/(2(-2)$

$=9\pm\sqrt{257}/(-4)$

$x=−8.75780488547035$ , $x=−0.7421951145296504$

There are no imaginary solutions.

How do you find the roots, real and imaginary, of y=-2(x +3)^2-7x+5y=-2(x +3)^2-7x+5 using the quadratic formula?