Solve Using the Quadratic Formula 4x^2-17x-15=0

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Algebra

$4 x^{2} - 17 x - 15 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 4$ , $b = - 17$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (4 \cdot - 15)}}{2 \cdot 4}$

Simplify.

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Simplify the numerator.

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Raise $- 17$ to the power of $2$ .

$x = \frac{17 \pm \sqrt{289 - 4 \cdot 4 \cdot - 15}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{17 \pm \sqrt{289 - 16 \cdot - 15}}{2 \cdot 4}$

Multiply $- 16$ by $- 15$ .

$x = \frac{17 \pm \sqrt{289 + 240}}{2 \cdot 4}$

Add $289$ and $240$ .

$x = \frac{17 \pm \sqrt{529}}{2 \cdot 4}$

Rewrite $529$ as $23^{2}$ .

$x = \frac{17 \pm \sqrt{23^{2}}}{2 \cdot 4}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{17 \pm 23}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{17 \pm 23}{8}$

The final answer is the combination of both solutions.

$x = 5, - \frac{3}{4}$

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