Solve Using the Quadratic Formula 4x^2+27x-7=0

$4 x^{2} + 27 x - 7 = 0$

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 $a = 4$ , $b = 27$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{- 27 \pm \sqrt{27^{2} - 4 \cdot (4 \cdot - 7)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $27$ to the power of $2$ .

$x = \frac{- 27 \pm \sqrt{729 - 4 \cdot 4 \cdot - 7}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 7$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 27 \pm \sqrt{729 - 16 \cdot - 7}}{2 \cdot 4}$

Multiply $- 16$ by $- 7$ .

$x = \frac{- 27 \pm \sqrt{729 + 112}}{2 \cdot 4}$

Add $729$ and $112$ .

$x = \frac{- 27 \pm \sqrt{841}}{2 \cdot 4}$

Rewrite $841$ as $29^{2}$ .

$x = \frac{- 27 \pm \sqrt{29^{2}}}{2 \cdot 4}$

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

$x = \frac{- 27 \pm 29}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 27 \pm 29}{8}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{1}{4}, - 7$