Solve Using the Quadratic Formula x^2-6=x

$x^{2} - 6 = x$

Step 1

Subtract $x$ from both sides of the equation.

$x^{2} - 6 - x = 0$

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 $a = 1$ , $b = - 1$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot - 6)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 6}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 6$ .

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot - 6}}{2 \cdot 1}$

Multiply $- 4$ by $- 6$ .

$x = \frac{1 \pm \sqrt{1 + 24}}{2 \cdot 1}$

Add $1$ and $24$ .

$x = \frac{1 \pm \sqrt{25}}{2 \cdot 1}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{1 \pm \sqrt{5^{2}}}{2 \cdot 1}$

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

$x = \frac{1 \pm 5}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm 5}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 3, - 2$