Solve x^2+20x+100=81

$${ x }^{ 2 } +20x+100=81$$

Solve for x

$$x=-19$$

$$x=-1$$

Steps Using Factoring

Subtract 81 from both sides.

$$x^{2}+20x+100-81=0$$

Subtract 81 from 100 to get 19.

$$x^{2}+20x+19=0$$

To solve the equation, factor x^{2}+20x+19 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.

$$a+b=20 ab=19$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=19$$

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

$$\left(x+1\right)\left(x+19\right)$$

To find equation solutions, solve x+1=0 and x+19=0.

$$x=-1 x=-19$$

Steps Using Factoring By Grouping

Subtract 81 from both sides.

$$x^{2}+20x+100-81=0$$

Subtract 81 from 100 to get 19.

$$x^{2}+20x+19=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+19. To find a and b, set up a system to be solved.

$$a+b=20 ab=1\times 19=19$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=19$$

Rewrite x^{2}+20x+19 as \left(x^{2}+x\right)+\left(19x+19\right).

$$\left(x^{2}+x\right)+\left(19x+19\right)$$

Factor out x in the first and 19 in the second group.

$$x\left(x+1\right)+19\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+19\right)$$

To find equation solutions, solve x+1=0 and x+19=0.

$$x=-1 x=-19$$

Steps Using the Quadratic Formula

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.

$$x^{2}+20x+100=81$$

Subtract 81 from both sides of the equation.

$$x^{2}+20x+100-81=81-81$$

Subtracting 81 from itself leaves 0.

$$x^{2}+20x+100-81=0$$

Subtract 81 from 100.

$$x^{2}+20x+19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and 19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 19}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 19}}{2}$$

Multiply -4 times 19.

$$x=\frac{-20±\sqrt{400-76}}{2}$$

Add 400 to -76.

$$x=\frac{-20±\sqrt{324}}{2}$$

Take the square root of 324.

$$x=\frac{-20±18}{2}$$

Now solve the equation x=\frac{-20±18}{2} when ± is plus. Add -20 to 18.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-20±18}{2} when ± is minus. Subtract 18 from -20.

$$x=-\frac{38}{2}$$

Divide -38 by 2.

$$x=-19$$

The equation is now solved.

$$x=-1 x=-19$$

Steps for Completing the Square

Factor x^{2}+20x+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}.

$$\left(x+10\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{81}$$

Simplify.

$$x+10=9 x+10=-9$$

Subtract 10 from both sides of the equation.

$$x=-1 x=-19$$