Solve 3x^2+3=9

$$3 x ^ { 2 } + 3 = 9$$

Solve for x

$$x=\sqrt{2}\approx 1.414213562$$

$$x=-\sqrt{2}\approx -1.414213562$$

Steps by Finding Square Root

Subtract 3 from both sides.

$$3x^{2}=9-3$$

Subtract 3 from 9 to get 6.

$$3x^{2}=6$$

Divide both sides by 3.

$$x^{2}=\frac{6}{3}$$

Divide 6 by 3 to get 2.

$$x^{2}=2$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Subtract 9 from both sides.

$$3x^{2}+3-9=0$$

Subtract 9 from 3 to get -6.

$$3x^{2}-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-6\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-6\right)}}{2\times 3}$$

Multiply -12 times -6.

$$x=\frac{0±\sqrt{72}}{2\times 3}$$

Take the square root of 72.

$$x=\frac{0±6\sqrt{2}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±6\sqrt{2}}{6}$$

Now solve the equation x=\frac{0±6\sqrt{2}}{6} when ± is plus.

$$x=\sqrt{2}$$

Now solve the equation x=\frac{0±6\sqrt{2}}{6} when ± is minus.

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

The equation is now solved.

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