Solve y+3=(x+4)^2

$$y + 3 = ( x + 4 ) ^ { 2 }$$

Solve for y

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

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$y+3=x^{2}+8x+16$$

Subtract 3 from both sides.

$$y=x^{2}+8x+16-3$$

Subtract 3 from 16 to get 13.

$$y=x^{2}+8x+13$$

Solve for x (complex solution)

$$x=-\left(\sqrt{y+3}+4\right)$$

$$x=\sqrt{y+3}-4$$

Solve for x

$$x=-\left(\sqrt{y+3}+4\right)$$

$$x=\sqrt{y+3}-4\text{, }y\geq -3$$