How do you find the zeros of $y = -2x^2 - 3x +12$ using the quadratic formula?

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Answer 1 · 2015-12-02T08:25:55

Set $a$ , $b$ , and $c$ appropriately and apply the quadratic formula to obtain the zeros at $x = (-3+sqrt(105))/4$ and $x = (-3-sqrt(105))/4$

The quadratic formula states that

$ax^2 + bx + c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)$

In this case, we have
$a = -2$
$b = -3$
$c = 12$

Thus the zeros, that is, the solutions to $-2x^2 - 3x + 12 = 0$ are

$x = (-(-3)+-sqrt((-3)^2-4(-2)(12)))/(2(-2))$

$=(3 +-sqrt(9 + 96))/(-4)$

$= -(3+-sqrt(105))/4$

$= (-3+-sqrt(105))/4$

So
$x = (-3+sqrt(105))/4$
or
$x = (-3-sqrt(105))/4$

How do you find the zeros of y = -2x^2 - 3x +12 y = -2x^2 - 3x +12 using the quadratic formula?