Solve Using the Quadratic Formula -4x^2-19x+5=0

$- 4 x^{2} - 19 x + 5 = 0$

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 $a = - 4$ , $b = - 19$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{19 \pm \sqrt{{(- 19)}^{2} - 4 \cdot (- 4 \cdot 5)}}{2 \cdot - 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 19$ to the power of $2$ .

$x = \frac{19 \pm \sqrt{361 - 4 \cdot - 4 \cdot 5}}{2 \cdot - 4}$

Multiply $- 4 \cdot - 4 \cdot 5$ .

Multiply $- 4$ by $- 4$ .

$x = \frac{19 \pm \sqrt{361 + 16 \cdot 5}}{2 \cdot - 4}$

Multiply $16$ by $5$ .

$x = \frac{19 \pm \sqrt{361 + 80}}{2 \cdot - 4}$

Add $361$ and $80$ .

$x = \frac{19 \pm \sqrt{441}}{2 \cdot - 4}$

Rewrite $441$ as $21^{2}$ .

$x = \frac{19 \pm \sqrt{21^{2}}}{2 \cdot - 4}$

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

$x = \frac{19 \pm 21}{2 \cdot - 4}$

Multiply $2$ by $- 4$ .

$x = \frac{19 \pm 21}{- 8}$

Move the negative in front of the fraction.

$x = - \frac{19 \pm 21}{8}$

Step 4

The final answer is the combination of both solutions.

$x = - 5, \frac{1}{4}$