Solve (3y+7)^2=3

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

Solve for y

$$y=\frac{\sqrt{3}-7}{3}\approx -1.755983064$$

$$y=\frac{-\sqrt{3}-7}{3}\approx -2.910683603$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

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

Subtract 7 from both sides of the equation.

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

Subtracting 7 from itself leaves 0.

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

Subtract 7 from \sqrt{3}.

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

Subtract 7 from -\sqrt{3}.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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