Solve (x-25)^2=100

$$( x - 25 ) ^ { 2 } = 100$$

Solve for x

$$x=35$$

$$x=15$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-25\right)^{2}.

$$x^{2}-50x+625=100$$

Subtract 100 from both sides.

$$x^{2}-50x+625-100=0$$

Subtract 100 from 625 to get 525.

$$x^{2}-50x+525=0$$

To solve the equation, factor x^{2}-50x+525 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-50 ab=525$$

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 525.

$$-1,-525 -3,-175 -5,-105 -7,-75 -15,-35 -21,-25$$

Calculate the sum for each pair.

$$-1-525=-526 -3-175=-178 -5-105=-110 -7-75=-82 -15-35=-50 -21-25=-46$$

The solution is the pair that gives sum -50.

$$a=-35 b=-15$$

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

$$\left(x-35\right)\left(x-15\right)$$

To find equation solutions, solve x-35=0 and x-15=0.

$$x=35 x=15$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-25\right)^{2}.

$$x^{2}-50x+625=100$$

Subtract 100 from both sides.

$$x^{2}-50x+625-100=0$$

Subtract 100 from 625 to get 525.

$$x^{2}-50x+525=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+525. To find a and b, set up a system to be solved.

$$a+b=-50 ab=1\times 525=525$$

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 525.

$$-1,-525 -3,-175 -5,-105 -7,-75 -15,-35 -21,-25$$

Calculate the sum for each pair.

$$-1-525=-526 -3-175=-178 -5-105=-110 -7-75=-82 -15-35=-50 -21-25=-46$$

The solution is the pair that gives sum -50.

$$a=-35 b=-15$$

Rewrite x^{2}-50x+525 as \left(x^{2}-35x\right)+\left(-15x+525\right).

$$\left(x^{2}-35x\right)+\left(-15x+525\right)$$

Factor out x in the first and -15 in the second group.

$$x\left(x-35\right)-15\left(x-35\right)$$

Factor out common term x-35 by using distributive property.

$$\left(x-35\right)\left(x-15\right)$$

To find equation solutions, solve x-35=0 and x-15=0.

$$x=35 x=15$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-25\right)^{2}.

$$x^{2}-50x+625=100$$

Subtract 100 from both sides.

$$x^{2}-50x+625-100=0$$

Subtract 100 from 625 to get 525.

$$x^{2}-50x+525=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -50 for b, and 525 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 525}}{2}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 525}}{2}$$

Multiply -4 times 525.

$$x=\frac{-\left(-50\right)±\sqrt{2500-2100}}{2}$$

Add 2500 to -2100.

$$x=\frac{-\left(-50\right)±\sqrt{400}}{2}$$

Take the square root of 400.

$$x=\frac{-\left(-50\right)±20}{2}$$

The opposite of -50 is 50.

$$x=\frac{50±20}{2}$$

Now solve the equation x=\frac{50±20}{2} when ± is plus. Add 50 to 20.

$$x=\frac{70}{2}$$

Divide 70 by 2.

$$x=35$$

Now solve the equation x=\frac{50±20}{2} when ± is minus. Subtract 20 from 50.

$$x=\frac{30}{2}$$

Divide 30 by 2.

$$x=15$$

The equation is now solved.

$$x=35 x=15$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x-25\right)^{2}}=\sqrt{100}$$

Simplify.

$$x-25=10 x-25=-10$$

Add 25 to both sides of the equation.

$$x=35 x=15$$