Solve Using the Quadratic Formula 7x^2-4x-3=0

$7 x^{2} - 4 x - 3 = 0$

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 $a = 7$ , $b = - 4$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 7 \cdot - 3}}{2 \cdot 7}$

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

Multiply $- 4$ by $7$ .

$x = \frac{4 \pm \sqrt{16 - 28 \cdot - 3}}{2 \cdot 7}$

Multiply $- 28$ by $- 3$ .

$x = \frac{4 \pm \sqrt{16 + 84}}{2 \cdot 7}$

Add $16$ and $84$ .

$x = \frac{4 \pm \sqrt{100}}{2 \cdot 7}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{4 \pm \sqrt{10^{2}}}{2 \cdot 7}$

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

$x = \frac{4 \pm 10}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{4 \pm 10}{14}$

Simplify $\frac{4 \pm 10}{14}$ .

$x = \frac{2 \pm 5}{7}$

Step 4

The final answer is the combination of both solutions.

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