Solve x^2=5^2+12^2

$${ x }^{ 2 } = { 5 }^{ 2 } + { 12 }^{ 2 }$$

Solve for x

$$x=-13$$

$$x=13$$

Steps Using Difference of Squares

Calculate 5 to the power of 2 and get 25.

$$x^{2}=25+12^{2}$$

Calculate 12 to the power of 2 and get 144.

$$x^{2}=25+144$$

Add 25 and 144 to get 169.

$$x^{2}=169$$

Subtract 169 from both sides.

$$x^{2}-169=0$$

Consider x^{2}-169. Rewrite x^{2}-169 as x^{2}-13^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-13\right)\left(x+13\right)=0$$

To find equation solutions, solve x-13=0 and x+13=0.

$$x=13 x=-13$$

Steps by Finding Square Root

Calculate 5 to the power of 2 and get 25.

$$x^{2}=25+12^{2}$$

Calculate 12 to the power of 2 and get 144.

$$x^{2}=25+144$$

Add 25 and 144 to get 169.

$$x^{2}=169$$

Take the square root of both sides of the equation.

$$x=13 x=-13$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$x^{2}=25+12^{2}$$

Calculate 12 to the power of 2 and get 144.

$$x^{2}=25+144$$

Add 25 and 144 to get 169.

$$x^{2}=169$$

Subtract 169 from both sides.

$$x^{2}-169=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -169 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-169\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-169\right)}}{2}$$

Multiply -4 times -169.

$$x=\frac{0±\sqrt{676}}{2}$$

Take the square root of 676.

$$x=\frac{0±26}{2}$$

Now solve the equation x=\frac{0±26}{2} when ± is plus. Divide 26 by 2.

$$x=13$$

Now solve the equation x=\frac{0±26}{2} when ± is minus. Divide -26 by 2.

$$x=-13$$

The equation is now solved.

$$x=13 x=-13$$