Solve Using the Quadratic Formula x-x^2+42=0

$x - x^{2} + 42 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 1$ , $b = 1$ , and $c = 42$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (- 1 \cdot 42)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 1 \cdot 42}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 \cdot 42}}{2 \cdot - 1}$

Multiply $4$ by $42$ .

$x = \frac{- 1 \pm \sqrt{1 + 168}}{2 \cdot - 1}$

Add $1$ and $168$ .

$x = \frac{- 1 \pm \sqrt{169}}{2 \cdot - 1}$

Rewrite $169$ as $13^{2}$ .

$x = \frac{- 1 \pm \sqrt{13^{2}}}{2 \cdot - 1}$

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

$x = \frac{- 1 \pm 13}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm 13}{- 2}$

Simplify $\frac{- 1 \pm 13}{- 2}$ .

$x = \frac{1 \pm 13}{2}$

The final answer is the combination of both solutions.

$x = 7, - 6$