Solve (sqrt{3-x^2})^2+(1+x)^2=13

$$( \sqrt { 3 - x ^ { 2 } } ) ^ { 2 } + ( 1 + x ) ^ { 2 } = 13$$

Solve for x (complex solution)

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps for Solving Linear Equation

Calculate \sqrt{3-x^{2}} to the power of 2 and get 3-x^{2}.

$$3-x^{2}+\left(1+x\right)^{2}=13$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$3-x^{2}+1+2x+x^{2}=13$$

Add 3 and 1 to get 4.

$$4-x^{2}+2x+x^{2}=13$$

Combine -x^{2} and x^{2} to get 0.

$$4+2x=13$$

Subtract 4 from both sides.

$$2x=13-4$$

Subtract 4 from 13 to get 9.

$$2x=9$$

Divide both sides by 2.

$$x=\frac{9}{2}$$