Find All Complex Solutions b^2-10b=39

$b^{2} - 10 b = 39$

Step 1

Subtract $39$ from both sides of the equation.

$b^{2} - 10 b - 39 = 0$

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 $a = 1$ , $b = - 10$ , and $c = - 39$ into the quadratic formula and solve for $b$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (1 \cdot - 39)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 10$ to the power of $2$ .

$b = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot - 39}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 39$ .

Multiply $- 4$ by $1$ .

$b = \frac{10 \pm \sqrt{100 - 4 \cdot - 39}}{2 \cdot 1}$

Multiply $- 4$ by $- 39$ .

$b = \frac{10 \pm \sqrt{100 + 156}}{2 \cdot 1}$

Add $100$ and $156$ .

$b = \frac{10 \pm \sqrt{256}}{2 \cdot 1}$

Rewrite $256$ as $16^{2}$ .

$b = \frac{10 \pm \sqrt{16^{2}}}{2 \cdot 1}$

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

$b = \frac{10 \pm 16}{2 \cdot 1}$

Multiply $2$ by $1$ .

$b = \frac{10 \pm 16}{2}$

Simplify $\frac{10 \pm 16}{2}$ .

$b = 5 \pm 8$

Step 5

The final answer is the combination of both solutions.

$b = 13, - 3$