Solve for x x^2+bx+16=0

$x^{2} + b x + 16 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = b$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- b \pm \sqrt{b^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Rewrite $4 \cdot 1 \cdot 16$ as ${(2 \cdot 1 \cdot 4)}^{2}$ .

$x = \frac{- b \pm \sqrt{b^{2} - {(2 \cdot 1 \cdot 4)}^{2}}}{2 \cdot 1}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 2 \cdot 1 \cdot 4$ .

$x = \frac{- b \pm \sqrt{(b + 2 \cdot 1 \cdot 4) (b - (2 \cdot 1 \cdot 4))}}{2 \cdot 1}$

Simplify.

Multiply $2 \cdot 1 \cdot 4$ .

Multiply $2$ by $1$ .

$x = \frac{- b \pm \sqrt{(b + 2 \cdot 4) (b - (2 \cdot 1 \cdot 4))}}{2 \cdot 1}$

Multiply $2$ by $4$ .

$x = \frac{- b \pm \sqrt{(b + 8) (b - (2 \cdot 1 \cdot 4))}}{2 \cdot 1}$

Multiply $- (2 \cdot 1 \cdot 4)$ .

Multiply $2$ by $1$ .

$x = \frac{- b \pm \sqrt{(b + 8) (b - (2 \cdot 4))}}{2 \cdot 1}$

Multiply $2$ by $4$ .

$x = \frac{- b \pm \sqrt{(b + 8) (b - 1 \cdot 8)}}{2 \cdot 1}$

Multiply $- 1$ by $8$ .

$x = \frac{- b \pm \sqrt{(b + 8) (b - 8)}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- b \pm \sqrt{(b + 8) (b - 8)}}{2}$

Step 4

The final answer is the combination of both solutions.

$x = - \frac{b - \sqrt{(b + 8) (b - 8)}}{2}$

$x = - \frac{b + \sqrt{(b + 8) (b - 8)}}{2}$

Solve for x x^2+bx+16=0

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