How do you solve $2 x^{2} + 3 x - 3 = 0$ $2x^2+3x-3=0$ using the quadratic formula?

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How do you solve $2 x^{2} + 3 x - 3 = 0$ $2x^2+3x-3=0$ using the quadratic formula?

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Answer 1 · 2016-03-17T11:18:59

$x = \frac{- 3 + \sqrt{33}}{4}$ $x= (-3+sqrt(33))/4$ or $x = \frac{- 3 - \sqrt{33}}{4}$ $x=(-3-sqrt(33))/4$

Explanation:

For a quadratic in the general standard form:
$XXX a x^{2} + b x + c = 0$ $color(white)("XXX")color(red)(a)x^2+color(blue)(b)x+color(green)(c)=0$
the quadratic formula for the solution(s) is
$XXX x = \frac{- b \pm \sqrt{b^{2} - 4 (a) (c)}}{2 (a)}$ $color(white)("XXX")x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4(color(red)(a))(color(green)(c))))/(2(color(red)(a)))$

For the given equation: $2 x^{2} + 3 x - 3$ $color(red)(2)x^2+color(blue)(3)xcolor(green)(-3)$
the quadratic formula becomes:
$XXX x = \frac{- 3 \pm \sqrt{3^{2} - 4 (2) (- 3)}}{2 (2)}$ $color(white)("XXX")x=(-color(blue)(3)+-sqrt(color(blue)(3)^2-4(color(red)(2))(color(green)(-3))))/(2(color(red)(2))$

$XXX x = \frac{- 3 \pm \sqrt{9 + 24}}{2 (2)}$ $color(white)("XXX")x=(-color(blue)(3)+-sqrt(color(blue)9+24))/(2(color(red)(2))$

$XXX = \frac{- 3 \pm \sqrt{33}}{4}$ $color(white)("XXX")=(-3+-sqrt(33))/4$

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How do you solve $2 x^{2} + 3 x - 3 = 0$ $2x^2+3x-3=0$ using the quadratic formula?

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