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 $4$ .

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

Step 4

The final answer is the combination of both solutions.

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