Solve sqrt{-20+6t}=t-2

$$\sqrt { - 20 + 6 t } = t - 2$$

Solve for t

$$t=6$$

$$t=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{-20+6t}\right)^{2}=\left(t-2\right)^{2}$$

Calculate \sqrt{-20+6t} to the power of 2 and get -20+6t.

$$-20+6t=\left(t-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-2\right)^{2}.

$$-20+6t=t^{2}-4t+4$$

Subtract t^{2} from both sides.

$$-20+6t-t^{2}=-4t+4$$

Add 4t to both sides.

$$-20+6t-t^{2}+4t=4$$

Combine 6t and 4t to get 10t.

$$-20+10t-t^{2}=4$$

Subtract 4 from both sides.

$$-20+10t-t^{2}-4=0$$

Subtract 4 from -20 to get -24.

$$-24+10t-t^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-t^{2}+10t-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt-24. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-24\right)=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 10.

$$a=6 b=4$$

Rewrite -t^{2}+10t-24 as \left(-t^{2}+6t\right)+\left(4t-24\right).

$$\left(-t^{2}+6t\right)+\left(4t-24\right)$$

Factor out -t in the first and 4 in the second group.

$$-t\left(t-6\right)+4\left(t-6\right)$$

Factor out common term t-6 by using distributive property.

$$\left(t-6\right)\left(-t+4\right)$$

To find equation solutions, solve t-6=0 and -t+4=0.

$$t=6 t=4$$

Substitute 6 for t in the equation \sqrt{-20+6t}=t-2.

$$\sqrt{-20+6\times 6}=6-2$$

Simplify. The value t=6 satisfies the equation.

$$4=4$$

Substitute 4 for t in the equation \sqrt{-20+6t}=t-2.

$$\sqrt{-20+6\times 4}=4-2$$

Simplify. The value t=4 satisfies the equation.

$$2=2$$

List all solutions of \sqrt{6t-20}=t-2.

$$t=6 t=4$$