Solve ((x)^(2)+2x)/(2)=6 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{ { x }^{ 2 } +2x }{ 2 } =6$$

Solve for x (complex solution)

$x=\sqrt{13}-1\approx 2.605551275$
$x=-\left(\sqrt{13}+1\right)\approx -4.605551275$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides by $2$.

$$x^{2}+2x=6\times 2$$

Multiply $6$ and $2$ to get $12$.

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

Subtract $12$ from both sides.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $2$ for $b$, and $-12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-12\right)}}{2}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\left(-12\right)}}{2}$$

Multiply $-4$ times $-12$.

$$x=\frac{-2±\sqrt{4+48}}{2}$$

Add $4$ to $48$.

$$x=\frac{-2±\sqrt{52}}{2}$$

Take the square root of $52$.

$$x=\frac{-2±2\sqrt{13}}{2}$$

Now solve the equation $x=\frac{-2±2\sqrt{13}}{2}$ when $±$ is plus. Add $-2$ to $2\sqrt{13}$.

$$x=\frac{2\sqrt{13}-2}{2}$$

Divide $-2+2\sqrt{13}$ by $2$.

$$x=\sqrt{13}-1$$

Now solve the equation $x=\frac{-2±2\sqrt{13}}{2}$ when $±$ is minus. Subtract $2\sqrt{13}$ from $-2$.

$$x=\frac{-2\sqrt{13}-2}{2}$$

Divide $-2-2\sqrt{13}$ by $2$.

$$x=-\sqrt{13}-1$$

The equation is now solved.

$$x=\sqrt{13}-1$$ $$x=-\sqrt{13}-1$$

Steps for Completing the Square

Multiply both sides by $2$.

$$x^{2}+2x=6\times 2$$

Multiply $6$ and $2$ to get $12$.

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

Divide $2$, the coefficient of the $x$ term, by $2$ to get $1$. Then add the square of $1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=12+1^{2}$$

Square $1$.

$$x^{2}+2x+1=12+1$$

Add $12$ to $1$.

$$x^{2}+2x+1=13$$

Factor $x^{2}+2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+1\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{13}$$

Simplify.

$$x+1=\sqrt{13}$$ $$x+1=-\sqrt{13}$$

Subtract $1$ from both sides of the equation.

$$x=\sqrt{13}-1$$ $$x=-\sqrt{13}-1$$

Solve for x

$x=\sqrt{13}-1\approx 2.605551275$
$x=-\sqrt{13}-1\approx -4.605551275$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides by $2$.

$$x^{2}+2x=6\times 2$$

Multiply $6$ and $2$ to get $12$.

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

Subtract $12$ from both sides.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $2$ for $b$, and $-12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-12\right)}}{2}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\left(-12\right)}}{2}$$

Multiply $-4$ times $-12$.

$$x=\frac{-2±\sqrt{4+48}}{2}$$

Add $4$ to $48$.

$$x=\frac{-2±\sqrt{52}}{2}$$

Take the square root of $52$.

$$x=\frac{-2±2\sqrt{13}}{2}$$

Now solve the equation $x=\frac{-2±2\sqrt{13}}{2}$ when $±$ is plus. Add $-2$ to $2\sqrt{13}$.

$$x=\frac{2\sqrt{13}-2}{2}$$

Divide $-2+2\sqrt{13}$ by $2$.

$$x=\sqrt{13}-1$$

Now solve the equation $x=\frac{-2±2\sqrt{13}}{2}$ when $±$ is minus. Subtract $2\sqrt{13}$ from $-2$.

$$x=\frac{-2\sqrt{13}-2}{2}$$

Divide $-2-2\sqrt{13}$ by $2$.

$$x=-\sqrt{13}-1$$

The equation is now solved.

$$x=\sqrt{13}-1$$ $$x=-\sqrt{13}-1$$

Steps for Completing the Square

Multiply both sides by $2$.

$$x^{2}+2x=6\times 2$$

Multiply $6$ and $2$ to get $12$.

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

Divide $2$, the coefficient of the $x$ term, by $2$ to get $1$. Then add the square of $1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=12+1^{2}$$

Square $1$.

$$x^{2}+2x+1=12+1$$

Add $12$ to $1$.

$$x^{2}+2x+1=13$$

Factor $x^{2}+2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+1\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{13}$$

Simplify.

$$x+1=\sqrt{13}$$ $$x+1=-\sqrt{13}$$

Subtract $1$ from both sides of the equation.

$$x=\sqrt{13}-1$$ $$x=-\sqrt{13}-1$$