Solve 9x^2+9=25

$$9 { x }^{ 2 } +9=25$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

Steps Using Difference of Squares

Subtract 25 from both sides.

$$9x^{2}+9-25=0$$

Subtract 25 from 9 to get -16.

$$9x^{2}-16=0$$

Consider 9x^{2}-16. Rewrite 9x^{2}-16 as \left(3x\right)^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(3x-4\right)\left(3x+4\right)=0$$

To find equation solutions, solve 3x-4=0 and 3x+4=0.

$$x=\frac{4}{3} x=-\frac{4}{3}$$

Steps by Finding Square Root

Subtract 9 from both sides.

$$9x^{2}=25-9$$

Subtract 9 from 25 to get 16.

$$9x^{2}=16$$

Divide both sides by 9.

$$x^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$x=\frac{4}{3} x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

Subtract 25 from both sides.

$$9x^{2}+9-25=0$$

Subtract 25 from 9 to get -16.

$$9x^{2}-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 9\left(-16\right)}}{2\times 9}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 9\left(-16\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{0±\sqrt{-36\left(-16\right)}}{2\times 9}$$

Multiply -36 times -16.

$$x=\frac{0±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$x=\frac{0±24}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{0±24}{18}$$

Now solve the equation x=\frac{0±24}{18} when ± is plus. Reduce the fraction \frac{24}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{0±24}{18} when ± is minus. Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{4}{3}$$

The equation is now solved.

$$x=\frac{4}{3} x=-\frac{4}{3}$$