Find All Complex Solutions 0=x^2+13x-30

$0 = x^{2} + 13 x - 30$

Step 1

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

Subtract $x^{2}$ from both sides of the equation.

$- x^{2} = 13 x - 30$

Subtract $13 x$ from both sides of the equation.

$- x^{2} - 13 x = - 30$

Add $30$ to both sides of the equation.

$- x^{2} - 13 x + 30 = 0$

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 $a = - 1$ , $b = - 13$ , and $c = 30$ into the quadratic formula and solve for $x$ .

$\frac{13 \pm \sqrt{{(- 13)}^{2} - 4 \cdot (- 1 \cdot 30)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 13$ to the power of $2$ .

$x = \frac{13 \pm \sqrt{169 - 4 \cdot - 1 \cdot 30}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 30$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{13 \pm \sqrt{169 + 4 \cdot 30}}{2 \cdot - 1}$

Multiply $4$ by $30$ .

$x = \frac{13 \pm \sqrt{169 + 120}}{2 \cdot - 1}$

Add $169$ and $120$ .

$x = \frac{13 \pm \sqrt{289}}{2 \cdot - 1}$

Rewrite $289$ as $17^{2}$ .

$x = \frac{13 \pm \sqrt{17^{2}}}{2 \cdot - 1}$

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

$x = \frac{13 \pm 17}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{13 \pm 17}{- 2}$

Move the negative in front of the fraction.

$x = - \frac{13 \pm 17}{2}$

Step 5

The final answer is the combination of both solutions.

$x = - 15, 2$