Solve -8(t-10)^2-8=64

$$- 8 ( t - 10 ) ^ { 2 } - 8 = 64$$

Solve for t

$$t=10-3i$$

$$t=10+3i$$

Steps by Finding Square Root

Subtracting 8 from itself leaves 0.

$$-8\left(t-10\right)^{2}=64+8$$

Add 64 to 8.

$$-8\left(t-10\right)^{2}=72$$

Divide both sides by -8.

$$\frac{-8\left(t-10\right)^{2}}{-8}=\frac{72}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$\left(t-10\right)^{2}=\frac{72}{-8}$$

Divide 72 by -8.

$$\left(t-10\right)^{2}=-9$$

Take the square root of both sides of the equation.

$$t-10=3i t-10=-3i$$

Add 10 to both sides of the equation.

$$t-10-\left(-10\right)=3i-\left(-10\right) t-10-\left(-10\right)=-3i-\left(-10\right)$$

Subtracting -10 from itself leaves 0.

$$t=3i-\left(-10\right) t=-3i-\left(-10\right)$$

Subtract -10 from 3i.

$$t=10+3i$$

Subtract -10 from -3i.

$$t=10-3i$$

The equation is now solved.

$$t=10+3i t=10-3i$$