Solve -40/9left(x-3right)^2+40=30

$$- \frac{ 40 }{ 9 } { \left(x-3 \right) }^{ 2 } +40 = 30$$

Solve for x

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

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps by Finding Square Root

Subtract 40 from both sides of the equation.

$$-\frac{40}{9}\left(x-3\right)^{2}+40-40=30-40$$

Subtracting 40 from itself leaves 0.

$$-\frac{40}{9}\left(x-3\right)^{2}=30-40$$

Subtract 40 from 30.

$$-\frac{40}{9}\left(x-3\right)^{2}=-10$$

Divide both sides of the equation by -\frac{40}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{40}{9}\left(x-3\right)^{2}}{-\frac{40}{9}}=-\frac{10}{-\frac{40}{9}}$$

Dividing by -\frac{40}{9} undoes the multiplication by -\frac{40}{9}.

$$\left(x-3\right)^{2}=-\frac{10}{-\frac{40}{9}}$$

Divide -10 by -\frac{40}{9} by multiplying -10 by the reciprocal of -\frac{40}{9}.

$$\left(x-3\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$x-3=\frac{3}{2} x-3=-\frac{3}{2}$$

Add 3 to both sides of the equation.

$$x-3-\left(-3\right)=\frac{3}{2}-\left(-3\right) x-3-\left(-3\right)=-\frac{3}{2}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$x=\frac{3}{2}-\left(-3\right) x=-\frac{3}{2}-\left(-3\right)$$

Subtract -3 from \frac{3}{2}.

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

Subtract -3 from -\frac{3}{2}.

$$x=\frac{3}{2}$$

The equation is now solved.

$$x=\frac{9}{2} x=\frac{3}{2}$$