Solve 9x^2+7=8

$$9 x ^ { 2 } + 7 = 8$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using Difference of Squares

Subtract 8 from both sides.

$$9x^{2}+7-8=0$$

Subtract 8 from 7 to get -1.

$$9x^{2}-1=0$$

Consider 9x^{2}-1. Rewrite 9x^{2}-1 as \left(3x\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(3x-1\right)\left(3x+1\right)=0$$

To find equation solutions, solve 3x-1=0 and 3x+1=0.

$$x=\frac{1}{3} x=-\frac{1}{3}$$

Steps by Finding Square Root

Subtract 7 from both sides.

$$9x^{2}=8-7$$

Subtract 7 from 8 to get 1.

$$9x^{2}=1$$

Divide both sides by 9.

$$x^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$x=\frac{1}{3} x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Subtract 8 from both sides.

$$9x^{2}+7-8=0$$

Subtract 8 from 7 to get -1.

$$9x^{2}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 9\left(-1\right)}}{2\times 9}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 9\left(-1\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{0±\sqrt{-36\left(-1\right)}}{2\times 9}$$

Multiply -36 times -1.

$$x=\frac{0±\sqrt{36}}{2\times 9}$$

Take the square root of 36.

$$x=\frac{0±6}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{0±6}{18}$$

Now solve the equation x=\frac{0±6}{18} when ± is plus. Reduce the fraction \frac{6}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{0±6}{18} when ± is minus. Reduce the fraction \frac{-6}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{3}$$

The equation is now solved.

$$x=\frac{1}{3} x=-\frac{1}{3}$$