Solve 2^2+9x^2=12

$$2 ^ { 2 } + 9 x ^ { 2 } = 12$$

Solve for x

$$x=\frac{2\sqrt{2}}{3}\approx 0.942809042$$

$$x=-\frac{2\sqrt{2}}{3}\approx -0.942809042$$

Steps by Finding Square Root

Calculate 2 to the power of 2 and get 4.

$$4+9x^{2}=12$$

Subtract 4 from both sides.

$$9x^{2}=12-4$$

Subtract 4 from 12 to get 8.

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

Divide both sides by 9.

$$x^{2}=\frac{8}{9}$$

Take the square root of both sides of the equation.

$$x=\frac{2\sqrt{2}}{3} x=-\frac{2\sqrt{2}}{3}$$

Steps Using the Quadratic Formula

Calculate 2 to the power of 2 and get 4.

$$4+9x^{2}=12$$

Subtract 12 from both sides.

$$4+9x^{2}-12=0$$

Subtract 12 from 4 to get -8.

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

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 9\left(-8\right)}}{2\times 9}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 9\left(-8\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{0±\sqrt{-36\left(-8\right)}}{2\times 9}$$

Multiply -36 times -8.

$$x=\frac{0±\sqrt{288}}{2\times 9}$$

Take the square root of 288.

$$x=\frac{0±12\sqrt{2}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{0±12\sqrt{2}}{18}$$

Now solve the equation x=\frac{0±12\sqrt{2}}{18} when ± is plus.

$$x=\frac{2\sqrt{2}}{3}$$

Now solve the equation x=\frac{0±12\sqrt{2}}{18} when ± is minus.

$$x=-\frac{2\sqrt{2}}{3}$$

The equation is now solved.

$$x=\frac{2\sqrt{2}}{3} x=-\frac{2\sqrt{2}}{3}$$