Solve {left(2sqrt{3}right)}^2+2^2=x^2

$${ \left(2 \sqrt{ 3 } \right) }^{ 2 } + { 2 }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x=4$$

$$x=-4$$

Steps Using Difference of Squares

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

The square of \sqrt{3} is 3.

$$4\times 3+2^{2}=x^{2}$$

Multiply 4 and 3 to get 12.

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

Calculate 2 to the power of 2 and get 4.

$$12+4=x^{2}$$

Add 12 and 4 to get 16.

$$16=x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}=16$$

Subtract 16 from both sides.

$$x^{2}-16=0$$

Consider x^{2}-16. Rewrite x^{2}-16 as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

To find equation solutions, solve x-4=0 and x+4=0.

$$x=4 x=-4$$

Steps by Finding Square Root

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

The square of \sqrt{3} is 3.

$$4\times 3+2^{2}=x^{2}$$

Multiply 4 and 3 to get 12.

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

Calculate 2 to the power of 2 and get 4.

$$12+4=x^{2}$$

Add 12 and 4 to get 16.

$$16=x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

Steps Using the Quadratic Formula

Expand \left(2\sqrt{3}\right)^{2}.

$$2^{2}\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

The square of \sqrt{3} is 3.

$$4\times 3+2^{2}=x^{2}$$

Multiply 4 and 3 to get 12.

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

Calculate 2 to the power of 2 and get 4.

$$12+4=x^{2}$$

Add 12 and 4 to get 16.

$$16=x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}=16$$

Subtract 16 from both sides.

$$x^{2}-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-16\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{0±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{0±8}{2}$$

Now solve the equation x=\frac{0±8}{2} when ± is plus. Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{0±8}{2} when ± is minus. Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=4 x=-4$$