How do you use the quadratic formula to solve $2 x^{2} - 5 x - 2 = 0$ $2x^2-5x-2=0$ ?

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How do you use the quadratic formula to solve $2 x^{2} - 5 x - 2 = 0$ $2x^2-5x-2=0$ ?

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Answer 1 · 2017-08-24T17:22:34

See a solution process below:

Explanation:

The quadratic formula states:

For $a x^{2} + b x + c = 0$ $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 \cdot a}$ $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$2$ $color(red)(2)$ for $a$ $color(red)(a)$

$- 5$ $color(blue)(-5)$ for $b$ $color(blue)(b)$

$- 2$ $color(green)(-2)$ for $c$ $color(green)(c)$ gives:

$x = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - (4 \cdot 2 \cdot - 2)}}{2 \cdot 2}$ $x = (-color(blue)((-5)) +- sqrt(color(blue)((-5))^2 - (4 * color(red)(2) * color(green)(-2))))/(2 * color(red)(2))$

$x = \frac{5 \pm \sqrt{25 - (- 16)}}{4}$ $x = (5 +- sqrt(25 - (-16)))/4$

$x = \frac{5 \pm \sqrt{25 + 16}}{4}$ $x = (5 +- sqrt(25 + 16))/4$

$x = \frac{5 \pm \sqrt{41}}{4}$ $x = (5 +- sqrt(41))/4$

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How do you use the quadratic formula to solve $2 x^{2} - 5 x - 2 = 0$ $2x^2-5x-2=0$ ?

1 Answer

Explanation:

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How do you use the quadratic formula to solve $2 x^{2} - 5 x - 2 = 0$ $2x^2-5x-2=0$ ?