Solve Using the Quadratic Formula x^2=-3x+40

$x^{2} = - 3 x + 40$

Step 1

Move all the expressions to the left side of the equation.

Add $3 x$ to both sides of the equation.

$x^{2} + 3 x = 40$

Subtract $40$ from both sides of the equation.

$x^{2} + 3 x - 40 = 0$

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 $a = 1$ , $b = 3$ , and $c = - 40$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 40)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 40}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 40$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 40}}{2 \cdot 1}$

Multiply $- 4$ by $- 40$ .

$x = \frac{- 3 \pm \sqrt{9 + 160}}{2 \cdot 1}$

Add $9$ and $160$ .

$x = \frac{- 3 \pm \sqrt{169}}{2 \cdot 1}$

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

$x = \frac{- 3 \pm \sqrt{13^{2}}}{2 \cdot 1}$

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

$x = \frac{- 3 \pm 13}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm 13}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 5, - 8$