Solve Using the Quadratic Formula 7x^2+6=13

$7 x^{2} + 6 = 13$

Step 1

Move all terms to the left side of the equation and simplify.

Subtract $13$ from both sides of the equation.

$7 x^{2} + 6 - 13 = 0$

Subtract $13$ from $6$ .

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

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 7$ , $b = 0$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (7 \cdot - 7)}}{2 \cdot 7}$

Step 4

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot 7 \cdot - 7}}{2 \cdot 7}$

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

Multiply $- 4$ by $7$ .

$x = \frac{0 \pm \sqrt{0 - 28 \cdot - 7}}{2 \cdot 7}$

Multiply $- 28$ by $- 7$ .

$x = \frac{0 \pm \sqrt{0 + 196}}{2 \cdot 7}$

Add $0$ and $196$ .

$x = \frac{0 \pm \sqrt{196}}{2 \cdot 7}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{0 \pm \sqrt{14^{2}}}{2 \cdot 7}$

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

$x = \frac{0 \pm 14}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{0 \pm 14}{14}$

Simplify $\frac{0 \pm 14}{14}$ .

$x = \pm 1$

Step 5

The final answer is the combination of both solutions.

$x = 1, - 1$