Solve Using the Quadratic Formula 7x^2-20x-3=0

$7 x^{2} - 20 x - 3 = 0$

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 $a = 7$ , $b = - 20$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{20 \pm \sqrt{{(- 20)}^{2} - 4 \cdot (7 \cdot - 3)}}{2 \cdot 7}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 20$ to the power of $2$ .

$x = \frac{20 \pm \sqrt{400 - 4 \cdot 7 \cdot - 3}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot - 3$ .

Multiply $- 4$ by $7$ .

$x = \frac{20 \pm \sqrt{400 - 28 \cdot - 3}}{2 \cdot 7}$

Multiply $- 28$ by $- 3$ .

$x = \frac{20 \pm \sqrt{400 + 84}}{2 \cdot 7}$

Add $400$ and $84$ .

$x = \frac{20 \pm \sqrt{484}}{2 \cdot 7}$

Rewrite $484$ as $22^{2}$ .

$x = \frac{20 \pm \sqrt{22^{2}}}{2 \cdot 7}$

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

$x = \frac{20 \pm 22}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{20 \pm 22}{14}$

Simplify $\frac{20 \pm 22}{14}$ .

$x = \frac{10 \pm 11}{7}$

Step 4

The final answer is the combination of both solutions.

$x = 3, - \frac{1}{7}$