Solve 17^2=x^2

$${ 17 }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x=-17$$

$$x=17$$

Steps Using Difference of Squares

Calculate 17 to the power of 2 and get 289.

$$289=x^{2}$$

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

$$x^{2}=289$$

Subtract 289 from both sides.

$$x^{2}-289=0$$

Consider x^{2}-289. Rewrite x^{2}-289 as x^{2}-17^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-17\right)\left(x+17\right)=0$$

To find equation solutions, solve x-17=0 and x+17=0.

$$x=17 x=-17$$

Steps by Finding Square Root

Calculate 17 to the power of 2 and get 289.

$$289=x^{2}$$

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

$$x^{2}=289$$

Take the square root of both sides of the equation.

$$x=17 x=-17$$

Steps Using the Quadratic Formula

Calculate 17 to the power of 2 and get 289.

$$289=x^{2}$$

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

$$x^{2}=289$$

Subtract 289 from both sides.

$$x^{2}-289=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -289 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-289\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-289\right)}}{2}$$

Multiply -4 times -289.

$$x=\frac{0±\sqrt{1156}}{2}$$

Take the square root of 1156.

$$x=\frac{0±34}{2}$$

Now solve the equation x=\frac{0±34}{2} when ± is plus. Divide 34 by 2.

$$x=17$$

Now solve the equation x=\frac{0±34}{2} when ± is minus. Divide -34 by 2.

$$x=-17$$

The equation is now solved.

$$x=17 x=-17$$