Solve 2x^2+17=179

$$2 { x }^{ 2 } +17=179$$

Solve for x

$$x=9$$

$$x=-9$$

Steps Using Difference of Squares

Subtract 179 from both sides.

$$2x^{2}+17-179=0$$

Subtract 179 from 17 to get -162.

$$2x^{2}-162=0$$

Divide both sides by 2.

$$x^{2}-81=0$$

Consider x^{2}-81. Rewrite x^{2}-81 as x^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-9\right)\left(x+9\right)=0$$

To find equation solutions, solve x-9=0 and x+9=0.

$$x=9 x=-9$$

Steps by Finding Square Root

Subtract 17 from both sides.

$$2x^{2}=179-17$$

Subtract 17 from 179 to get 162.

$$2x^{2}=162$$

Divide both sides by 2.

$$x^{2}=\frac{162}{2}$$

Divide 162 by 2 to get 81.

$$x^{2}=81$$

Take the square root of both sides of the equation.

$$x=9 x=-9$$

Steps Using the Quadratic Formula

Subtract 179 from both sides.

$$2x^{2}+17-179=0$$

Subtract 179 from 17 to get -162.

$$2x^{2}-162=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -162 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-162\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-162\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-162\right)}}{2\times 2}$$

Multiply -8 times -162.

$$x=\frac{0±\sqrt{1296}}{2\times 2}$$

Take the square root of 1296.

$$x=\frac{0±36}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±36}{4}$$

Now solve the equation x=\frac{0±36}{4} when ± is plus. Divide 36 by 4.

$$x=9$$

Now solve the equation x=\frac{0±36}{4} when ± is minus. Divide -36 by 4.

$$x=-9$$

The equation is now solved.

$$x=9 x=-9$$