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

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

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

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

Step 3

Simplify.

Simplify the numerator.

Raise $- 15$ to the power of $2$ .

$x = \frac{15 \pm \sqrt{225 - 4 \cdot - 4 \cdot 4}}{2 \cdot - 4}$

Multiply $- 4 \cdot - 4 \cdot 4$ .

Multiply $- 4$ by $- 4$ .

$x = \frac{15 \pm \sqrt{225 + 16 \cdot 4}}{2 \cdot - 4}$

Multiply $16$ by $4$ .

$x = \frac{15 \pm \sqrt{225 + 64}}{2 \cdot - 4}$

Add $225$ and $64$ .

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

Rewrite $289$ as $17^{2}$ .

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

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

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

Multiply $2$ by $- 4$ .

$x = \frac{15 \pm 17}{- 8}$

Move the negative in front of the fraction.

$x = - \frac{15 \pm 17}{8}$

Step 4

The final answer is the combination of both solutions.

$x = - 4, \frac{1}{4}$