Find All Complex Solutions 18=6x+x(x-13)

$18 = 6 x + x (x - 13)$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the right side.

Simplify $6 x + x (x - 13)$ .

Simplify each term.

Apply the distributive property.

$18 = 6 x + x \cdot x + x \cdot - 13$

Multiply $x$ by $x$ .

$18 = 6 x + x^{2} + x \cdot - 13$

Move $- 13$ to the left of $x$ .

$18 = 6 x + x^{2} - 13 x$

Subtract $13 x$ from $6 x$ .

$18 = x^{2} - 7 x$

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

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

$18 - x^{2} = - 7 x$

Add $7 x$ to both sides of the equation.

$18 - x^{2} + 7 x = 0$

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 $a = - 1$ , $b = 7$ , and $c = 18$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 1 \cdot 18)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 1 \cdot 18}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 18$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{- 7 \pm \sqrt{49 + 4 \cdot 18}}{2 \cdot - 1}$

Multiply $4$ by $18$ .

$x = \frac{- 7 \pm \sqrt{49 + 72}}{2 \cdot - 1}$

Add $49$ and $72$ .

$x = \frac{- 7 \pm \sqrt{121}}{2 \cdot - 1}$

Rewrite $121$ as $11^{2}$ .

$x = \frac{- 7 \pm \sqrt{11^{2}}}{2 \cdot - 1}$

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

$x = \frac{- 7 \pm 11}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 7 \pm 11}{- 2}$

Simplify $\frac{- 7 \pm 11}{- 2}$ .

$x = \frac{7 \pm 11}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 9, - 2$

Find All Complex Solutions 18=6x+x(x-13)

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