Solve for c x^2+18x+c=25+c

$x^{2} + 18 x + c = 25 + c$

Move all terms containing $c$ to the left side of the equation.

Subtract $c$ from both sides of the equation.

$x^{2} + 18 x + c - c = 25$

Combine the opposite terms in $x^{2} + 18 x + c - c$ .

Subtract $c$ from $c$ .

$x^{2} + 18 x + 0 = 25$

Add $x^{2} + 18 x$ and $0$ .

$x^{2} + 18 x = 25$

Subtract $25$ from both sides of the equation.

$x^{2} + 18 x - 25 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 18$ , and $c = - 25$ into the quadratic formula and solve for $x$ .

$\frac{- 18 \pm \sqrt{18^{2} - 4 \cdot (1 \cdot - 25)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $18$ to the power of $2$ .

$x = \frac{- 18 \pm \sqrt{324 - 4 \cdot 1 \cdot - 25}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 18 \pm \sqrt{324 - 4 \cdot - 25}}{2 \cdot 1}$

Multiply $- 4$ by $- 25$ .

$x = \frac{- 18 \pm \sqrt{324 + 100}}{2 \cdot 1}$

Add $324$ and $100$ .

$x = \frac{- 18 \pm \sqrt{424}}{2 \cdot 1}$

Rewrite $424$ as $2^{2} \cdot 106$ .

Factor $4$ out of $424$ .

$x = \frac{- 18 \pm \sqrt{4 (106)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 18 \pm \sqrt{2^{2} \cdot 106}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 18 \pm 2 \sqrt{106}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 18 \pm 2 \sqrt{106}}{2}$

Simplify $\frac{- 18 \pm 2 \sqrt{106}}{2}$ .

$x = - 9 \pm \sqrt{106}$

The final answer is the combination of both solutions.

$x = - 9 + \sqrt{106}, - 9 - \sqrt{106}$

The variable $c$ got canceled.

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$