Solve 3x^2+40=12

$$3 x ^ { 2 } + 40 = 12$$

Solve for x (complex solution)

$$x=-\frac{2\sqrt{21}i}{3}\approx -0-3.055050463i$$

$$x=\frac{2\sqrt{21}i}{3}\approx 3.055050463i$$

Steps by Finding Square Root

Subtract 40 from both sides.

$$3x^{2}=12-40$$

Subtract 40 from 12 to get -28.

$$3x^{2}=-28$$

Divide both sides by 3.

$$x^{2}=-\frac{28}{3}$$

The equation is now solved.

$$x=\frac{2\sqrt{21}i}{3} x=-\frac{2\sqrt{21}i}{3}$$

Steps Using the Quadratic Formula

Subtract 12 from both sides.

$$3x^{2}+40-12=0$$

Subtract 12 from 40 to get 28.

$$3x^{2}+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\times 28}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\times 28}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\times 28}}{2\times 3}$$

Multiply -12 times 28.

$$x=\frac{0±\sqrt{-336}}{2\times 3}$$

Take the square root of -336.

$$x=\frac{0±4\sqrt{21}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±4\sqrt{21}i}{6}$$

Now solve the equation x=\frac{0±4\sqrt{21}i}{6} when ± is plus.

$$x=\frac{2\sqrt{21}i}{3}$$

Now solve the equation x=\frac{0±4\sqrt{21}i}{6} when ± is minus.

$$x=-\frac{2\sqrt{21}i}{3}$$

The equation is now solved.

$$x=\frac{2\sqrt{21}i}{3} x=-\frac{2\sqrt{21}i}{3}$$