Solve 30^2+40^2=(x+12)^2

$$30 ^ { 2 } + 40 ^ { 2 } = ( x + 12 ) ^ { 2 }$$

Solve for x

$$x=38$$

$$x=-62$$

Steps Using Factoring

Calculate 30 to the power of 2 and get 900.

$$900+40^{2}=\left(x+12\right)^{2}$$

Calculate 40 to the power of 2 and get 1600.

$$900+1600=\left(x+12\right)^{2}$$

Add 900 and 1600 to get 2500.

$$2500=\left(x+12\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+12\right)^{2}.

$$2500=x^{2}+24x+144$$

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

$$x^{2}+24x+144=2500$$

Subtract 2500 from both sides.

$$x^{2}+24x+144-2500=0$$

Subtract 2500 from 144 to get -2356.

$$x^{2}+24x-2356=0$$

To solve the equation, factor x^{2}+24x-2356 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.

$$a+b=24 ab=-2356$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -2356.

$$-1,2356 -2,1178 -4,589 -19,124 -31,76 -38,62$$

Calculate the sum for each pair.

$$-1+2356=2355 -2+1178=1176 -4+589=585 -19+124=105 -31+76=45 -38+62=24$$

The solution is the pair that gives sum 24.

$$a=-38 b=62$$

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

$$\left(x-38\right)\left(x+62\right)$$

To find equation solutions, solve x-38=0 and x+62=0.

$$x=38 x=-62$$

Steps Using Factoring By Grouping

Calculate 30 to the power of 2 and get 900.

$$900+40^{2}=\left(x+12\right)^{2}$$

Calculate 40 to the power of 2 and get 1600.

$$900+1600=\left(x+12\right)^{2}$$

Add 900 and 1600 to get 2500.

$$2500=\left(x+12\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+12\right)^{2}.

$$2500=x^{2}+24x+144$$

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

$$x^{2}+24x+144=2500$$

Subtract 2500 from both sides.

$$x^{2}+24x+144-2500=0$$

Subtract 2500 from 144 to get -2356.

$$x^{2}+24x-2356=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-2356. To find a and b, set up a system to be solved.

$$a+b=24 ab=1\left(-2356\right)=-2356$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -2356.

$$-1,2356 -2,1178 -4,589 -19,124 -31,76 -38,62$$

Calculate the sum for each pair.

$$-1+2356=2355 -2+1178=1176 -4+589=585 -19+124=105 -31+76=45 -38+62=24$$

The solution is the pair that gives sum 24.

$$a=-38 b=62$$

Rewrite x^{2}+24x-2356 as \left(x^{2}-38x\right)+\left(62x-2356\right).

$$\left(x^{2}-38x\right)+\left(62x-2356\right)$$

Factor out x in the first and 62 in the second group.

$$x\left(x-38\right)+62\left(x-38\right)$$

Factor out common term x-38 by using distributive property.

$$\left(x-38\right)\left(x+62\right)$$

To find equation solutions, solve x-38=0 and x+62=0.

$$x=38 x=-62$$

Steps Using the Quadratic Formula

Calculate 30 to the power of 2 and get 900.

$$900+40^{2}=\left(x+12\right)^{2}$$

Calculate 40 to the power of 2 and get 1600.

$$900+1600=\left(x+12\right)^{2}$$

Add 900 and 1600 to get 2500.

$$2500=\left(x+12\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+12\right)^{2}.

$$2500=x^{2}+24x+144$$

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

$$x^{2}+24x+144=2500$$

Subtract 2500 from both sides.

$$x^{2}+24x+144-2500=0$$

Subtract 2500 from 144 to get -2356.

$$x^{2}+24x-2356=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 24 for b, and -2356 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\left(-2356\right)}}{2}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\left(-2356\right)}}{2}$$

Multiply -4 times -2356.

$$x=\frac{-24±\sqrt{576+9424}}{2}$$

Add 576 to 9424.

$$x=\frac{-24±\sqrt{10000}}{2}$$

Take the square root of 10000.

$$x=\frac{-24±100}{2}$$

Now solve the equation x=\frac{-24±100}{2} when ± is plus. Add -24 to 100.

$$x=\frac{76}{2}$$

Divide 76 by 2.

$$x=38$$

Now solve the equation x=\frac{-24±100}{2} when ± is minus. Subtract 100 from -24.

$$x=-\frac{124}{2}$$

Divide -124 by 2.

$$x=-62$$

The equation is now solved.

$$x=38 x=-62$$

Steps for Completing the Square

Calculate 30 to the power of 2 and get 900.

$$900+40^{2}=\left(x+12\right)^{2}$$

Calculate 40 to the power of 2 and get 1600.

$$900+1600=\left(x+12\right)^{2}$$

Add 900 and 1600 to get 2500.

$$2500=\left(x+12\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+12\right)^{2}.

$$2500=x^{2}+24x+144$$

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

$$x^{2}+24x+144=2500$$

Factor x^{2}+24x+144. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+12\right)^{2}=2500$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+12\right)^{2}}=\sqrt{2500}$$

Simplify.

$$x+12=50 x+12=-50$$

Subtract 12 from both sides of the equation.

$$x=38 x=-62$$