Solve for x 2(6-x)=x(x+5)

$2 (6 - x) = x (x + 5)$

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x (x + 5) = 2 (6 - x)$

Simplify $x (x + 5)$ .

Start simplifying.

$0 + 0 + x (x + 5) = 2 (6 - x)$

Simplify by adding zeros.

$x (x + 5) = 2 (6 - x)$

Apply the distributive property.

$x \cdot x + x \cdot 5 = 2 (6 - x)$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 5 = 2 (6 - x)$

Move $5$ to the left of $x$ .

$x^{2} + 5 x = 2 (6 - x)$

Simplify $2 (6 - x)$ .

Apply the distributive property.

$x^{2} + 5 x = 2 \cdot 6 + 2 (- x)$

Multiply.

Multiply $2$ by $6$ .

$x^{2} + 5 x = 12 + 2 (- x)$

Multiply $- 1$ by $2$ .

$x^{2} + 5 x = 12 - 2 x$

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

Add $2 x$ to both sides of the equation.

$x^{2} + 5 x + 2 x = 12$

Add $5 x$ and $2 x$ .

$x^{2} + 7 x = 12$

Subtract $12$ from both sides of the equation.

$x^{2} + 7 x - 12 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 7$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (1 \cdot - 12)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 12}}{2 \cdot 1}$

Multiply $- 4$ by $- 12$ .

$x = \frac{- 7 \pm \sqrt{49 + 48}}{2 \cdot 1}$

Add $49$ and $48$ .

$x = \frac{- 7 \pm \sqrt{97}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 7 \pm \sqrt{97}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{7 - \sqrt{97}}{2}, - \frac{7 + \sqrt{97}}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{7 - \sqrt{97}}{2}, - \frac{7 + \sqrt{97}}{2}$

Decimal Form:

$x = 1.42442890 \dots, - 8.42442890 \dots$

Solve for x 2(6-x)=x(x+5)

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