Solve (2y+3)^2-7=0

$$( 2 y + 3 ) ^ { 2 } - 7 = 0$$

Solve for y

$$y=\frac{\sqrt{7}-3}{2}\approx -0.177124344$$

$$y=\frac{-\sqrt{7}-3}{2}\approx -2.822875656$$

Steps by Finding Square Root

Add 7 to both sides of the equation.

$$\left(2y+3\right)^{2}-7+7=7$$

Subtracting 7 from itself leaves 0.

$$\left(2y+3\right)^{2}=7$$

Take the square root of both sides of the equation.

$$2y+3=\sqrt{7} 2y+3=-\sqrt{7}$$

Subtract 3 from both sides of the equation.

$$2y+3-3=\sqrt{7}-3 2y+3-3=-\sqrt{7}-3$$

Subtracting 3 from itself leaves 0.

$$2y=\sqrt{7}-3 2y=-\sqrt{7}-3$$

Subtract 3 from \sqrt{7}.

$$2y=\sqrt{7}-3$$

Subtract 3 from -\sqrt{7}.

$$2y=-\sqrt{7}-3$$

Divide both sides by 2.

$$\frac{2y}{2}=\frac{\sqrt{7}-3}{2} \frac{2y}{2}=\frac{-\sqrt{7}-3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y=\frac{\sqrt{7}-3}{2} y=\frac{-\sqrt{7}-3}{2}$$