Solve 5^2=4^2+(10-t)^2

$$5 ^ { 2 } = 4 ^ { 2 } + ( 10 - t ) ^ { 2 }$$

Solve for t

$$t=7$$

$$t=13$$

Steps Using Factoring

Calculate 5 to the power of 2 and get 25.

$$25=4^{2}+\left(10-t\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$25=16+\left(10-t\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-t\right)^{2}.

$$25=16+100-20t+t^{2}$$

Add 16 and 100 to get 116.

$$25=116-20t+t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$116-20t+t^{2}=25$$

Subtract 25 from both sides.

$$116-20t+t^{2}-25=0$$

Subtract 25 from 116 to get 91.

$$91-20t+t^{2}=0$$

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

$$t^{2}-20t+91=0$$

To solve the equation, factor t^{2}-20t+91 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=-20 ab=91$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 91.

$$-1,-91 -7,-13$$

Calculate the sum for each pair.

$$-1-91=-92 -7-13=-20$$

The solution is the pair that gives sum -20.

$$a=-13 b=-7$$

Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.

$$\left(t-13\right)\left(t-7\right)$$

To find equation solutions, solve t-13=0 and t-7=0.

$$t=13 t=7$$

Steps Using Factoring By Grouping

Calculate 5 to the power of 2 and get 25.

$$25=4^{2}+\left(10-t\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$25=16+\left(10-t\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-t\right)^{2}.

$$25=16+100-20t+t^{2}$$

Add 16 and 100 to get 116.

$$25=116-20t+t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$116-20t+t^{2}=25$$

Subtract 25 from both sides.

$$116-20t+t^{2}-25=0$$

Subtract 25 from 116 to get 91.

$$91-20t+t^{2}=0$$

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

$$t^{2}-20t+91=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+91. To find a and b, set up a system to be solved.

$$a+b=-20 ab=1\times 91=91$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 91.

$$-1,-91 -7,-13$$

Calculate the sum for each pair.

$$-1-91=-92 -7-13=-20$$

The solution is the pair that gives sum -20.

$$a=-13 b=-7$$

Rewrite t^{2}-20t+91 as \left(t^{2}-13t\right)+\left(-7t+91\right).

$$\left(t^{2}-13t\right)+\left(-7t+91\right)$$

Factor out t in the first and -7 in the second group.

$$t\left(t-13\right)-7\left(t-13\right)$$

Factor out common term t-13 by using distributive property.

$$\left(t-13\right)\left(t-7\right)$$

To find equation solutions, solve t-13=0 and t-7=0.

$$t=13 t=7$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$25=4^{2}+\left(10-t\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$25=16+\left(10-t\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-t\right)^{2}.

$$25=16+100-20t+t^{2}$$

Add 16 and 100 to get 116.

$$25=116-20t+t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$116-20t+t^{2}=25$$

Subtract 25 from both sides.

$$116-20t+t^{2}-25=0$$

Subtract 25 from 116 to get 91.

$$91-20t+t^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$t^{2}-20t+91=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -20 for b, and 91 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 91}}{2}$$

Square -20.

$$t=\frac{-\left(-20\right)±\sqrt{400-4\times 91}}{2}$$

Multiply -4 times 91.

$$t=\frac{-\left(-20\right)±\sqrt{400-364}}{2}$$

Add 400 to -364.

$$t=\frac{-\left(-20\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$t=\frac{-\left(-20\right)±6}{2}$$

The opposite of -20 is 20.

$$t=\frac{20±6}{2}$$

Now solve the equation t=\frac{20±6}{2} when ± is plus. Add 20 to 6.

$$t=\frac{26}{2}$$

Divide 26 by 2.

$$t=13$$

Now solve the equation t=\frac{20±6}{2} when ± is minus. Subtract 6 from 20.

$$t=\frac{14}{2}$$

Divide 14 by 2.

$$t=7$$

The equation is now solved.

$$t=13 t=7$$

Steps for Completing the Square

Calculate 5 to the power of 2 and get 25.

$$25=4^{2}+\left(10-t\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$25=16+\left(10-t\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-t\right)^{2}.

$$25=16+100-20t+t^{2}$$

Add 16 and 100 to get 116.

$$25=116-20t+t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$116-20t+t^{2}=25$$

Subtract 116 from both sides.

$$-20t+t^{2}=25-116$$

Subtract 116 from 25 to get -91.

$$-20t+t^{2}=-91$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$t^{2}-20t=-91$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-20t+\left(-10\right)^{2}=-91+\left(-10\right)^{2}$$

Square -10.

$$t^{2}-20t+100=-91+100$$

Add -91 to 100.

$$t^{2}-20t+100=9$$

Factor t^{2}-20t+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 10 to both sides of the equation.

$$t=13 t=7$$