How to solve (x)^(2)+6=18

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (x)^(2)+6=18 | Equatix

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Question

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

Solve for x

$x=2\sqrt{3}\approx 3.464101615$
$x=-2\sqrt{3}\approx -3.464101615$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Subtract $6$ from both sides.

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

Subtract $6$ from $18$ to get $12$.

$$x^{2}=12$$

Take the square root of both sides of the equation.

$$x=2\sqrt{3}$$ $$x=-2\sqrt{3}$$

Steps Using the Quadratic Formula

Subtract $18$ from both sides.

$$x^{2}+6-18=0$$

Subtract $18$ from $6$ to get $-12$.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-12\right)}}{2}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\left(-12\right)}}{2}$$

Multiply $-4$ times $-12$.

$$x=\frac{0±\sqrt{48}}{2}$$

Take the square root of $48$.

$$x=\frac{0±4\sqrt{3}}{2}$$

Now solve the equation $x=\frac{0±4\sqrt{3}}{2}$ when $±$ is plus.

$$x=2\sqrt{3}$$

Now solve the equation $x=\frac{0±4\sqrt{3}}{2}$ when $±$ is minus.

$$x=-2\sqrt{3}$$

The equation is now solved.

$$x=2\sqrt{3}$$ $$x=-2\sqrt{3}$$

Show Solution

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Example

Question

Solve for x

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Steps Using the Quadratic Formula