Solve 10(1-t)^2=5

$$10 ( 1 - t ) ^ { 2 } = 5$$

Solve for t

$$t=-\frac{\sqrt{2}}{2}+1\approx 0.292893219$$

$$t=\frac{\sqrt{2}}{2}+1\approx 1.707106781$$

Steps by Finding Square Root

Divide both sides by 10.

$$\frac{10\left(-t+1\right)^{2}}{10}=\frac{5}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$\left(-t+1\right)^{2}=\frac{5}{10}$$

Reduce the fraction \frac{5}{10} to lowest terms by extracting and canceling out 5.

$$\left(-t+1\right)^{2}=\frac{1}{2}$$

Take the square root of both sides of the equation.

$$-t+1=\frac{\sqrt{2}}{2} -t+1=-\frac{\sqrt{2}}{2}$$

Subtract 1 from both sides of the equation.

$$-t+1-1=\frac{\sqrt{2}}{2}-1 -t+1-1=-\frac{\sqrt{2}}{2}-1$$

Subtracting 1 from itself leaves 0.

$$-t=\frac{\sqrt{2}}{2}-1 -t=-\frac{\sqrt{2}}{2}-1$$

Subtract 1 from \frac{\sqrt{2}}{2}.

$$-t=\frac{\sqrt{2}}{2}-1$$

Subtract 1 from -\frac{\sqrt{2}}{2}.

$$-t=-\frac{\sqrt{2}}{2}-1$$

Divide both sides by -1.

$$\frac{-t}{-1}=\frac{\frac{\sqrt{2}}{2}-1}{-1} \frac{-t}{-1}=\frac{-\frac{\sqrt{2}}{2}-1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t=\frac{\frac{\sqrt{2}}{2}-1}{-1} t=\frac{-\frac{\sqrt{2}}{2}-1}{-1}$$

Divide \frac{\sqrt{2}}{2}-1 by -1.

$$t=-\frac{\sqrt{2}}{2}+1$$

Divide -\frac{\sqrt{2}}{2}-1 by -1.

$$t=\frac{\sqrt{2}}{2}+1$$

The equation is now solved.

$$t=-\frac{\sqrt{2}}{2}+1 t=\frac{\sqrt{2}}{2}+1$$