How do you solve $2 x^{2} + 9 x - 18 = 0$ $2x^2+9x-18=0$ ?

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How do you solve $2 x^{2} + 9 x - 18 = 0$ $2x^2+9x-18=0$ ?

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Answer 1 · 2017-02-08T21:35:13

$x = \frac{3}{2}$ $x = 3/2$ or $x = - 6$ $x = -6$

Explanation:

$2 x^{2} + 9 x - 18 = 0$ $2x^2+9x-18=0$

use the ABC Formula, where you use:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b ± sqrt(b^2 - 4ac))/(2a)$
with:
$a = 2$ $a=2$
$b = 9$ $b=9$
$c = - 18$ $c=-18$

so when you fill that in:

$x = \frac{- 9 + \sqrt{9^{2} - (4 \cdot 2 \cdot - 18)}}{4} = \frac{3}{2}$ $x = (-9 + sqrt(9^2 - (4*2*-18)))/(4) = 3/2$
or
$x = \frac{- 9 - \sqrt{9^{2} - (4 \cdot 2 \cdot - 18)}}{4} = - 6$ $x = (-9 - sqrt(9^2 - (4*2*-18)))/(4) = -6$

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How do you solve $2 x^{2} + 9 x - 18 = 0$ $2x^2+9x-18=0$ ?

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