Solve x^2-25=(x-5)

$$x ^ { 2 } - 25 = ( x - 5 )$$

Solve for x

$$x=-4$$

$$x=5$$

Steps Using Factoring

Subtract x from both sides.

$$x^{2}-25-x=-5$$

Add 5 to both sides.

$$x^{2}-25-x+5=0$$

Add -25 and 5 to get -20.

$$x^{2}-20-x=0$$

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

$$x^{2}-x-20=0$$

To solve the equation, factor x^{2}-x-20 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.

$$a+b=-1 ab=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

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

$$\left(x-5\right)\left(x+4\right)$$

To find equation solutions, solve x-5=0 and x+4=0.

$$x=5 x=-4$$

Steps Using Factoring By Grouping

Subtract x from both sides.

$$x^{2}-25-x=-5$$

Add 5 to both sides.

$$x^{2}-25-x+5=0$$

Add -25 and 5 to get -20.

$$x^{2}-20-x=0$$

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

$$x^{2}-x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-20\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

Rewrite x^{2}-x-20 as \left(x^{2}-5x\right)+\left(4x-20\right).

$$\left(x^{2}-5x\right)+\left(4x-20\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x-5\right)+4\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+4\right)$$

To find equation solutions, solve x-5=0 and x+4=0.

$$x=5 x=-4$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$x^{2}-25-x=-5$$

Add 5 to both sides.

$$x^{2}-25-x+5=0$$

Add -25 and 5 to get -20.

$$x^{2}-20-x=0$$

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.

$$x^{2}-x-20=0$$

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

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-20\right)}}{2}$$

Multiply -4 times -20.

$$x=\frac{-\left(-1\right)±\sqrt{1+80}}{2}$$

Add 1 to 80.

$$x=\frac{-\left(-1\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-\left(-1\right)±9}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±9}{2}$$

Now solve the equation x=\frac{1±9}{2} when ± is plus. Add 1 to 9.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{1±9}{2} when ± is minus. Subtract 9 from 1.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=5 x=-4$$

Steps for Completing the Square

Subtract x from both sides.

$$x^{2}-25-x=-5$$

Add 25 to both sides.

$$x^{2}-x=-5+25$$

Add -5 and 25 to get 20.

$$x^{2}-x=20$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=20+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=20+\frac{1}{4}$$

Add 20 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{81}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{9}{2} x-\frac{1}{2}=-\frac{9}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=5 x=-4$$