Solve 3x^2+4=436

$$3 x ^ { 2 } + 4 = 436$$

Solve for x

$$x=12$$

$$x=-12$$

Steps Using Difference of Squares

Subtract 436 from both sides.

$$3x^{2}+4-436=0$$

Subtract 436 from 4 to get -432.

$$3x^{2}-432=0$$

Divide both sides by 3.

$$x^{2}-144=0$$

Consider x^{2}-144. Rewrite x^{2}-144 as x^{2}-12^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-12\right)\left(x+12\right)=0$$

To find equation solutions, solve x-12=0 and x+12=0.

$$x=12 x=-12$$

Steps by Finding Square Root

Subtract 4 from both sides.

$$3x^{2}=436-4$$

Subtract 4 from 436 to get 432.

$$3x^{2}=432$$

Divide both sides by 3.

$$x^{2}=\frac{432}{3}$$

Divide 432 by 3 to get 144.

$$x^{2}=144$$

Take the square root of both sides of the equation.

$$x=12 x=-12$$

Steps Using the Quadratic Formula

Subtract 436 from both sides.

$$3x^{2}+4-436=0$$

Subtract 436 from 4 to get -432.

$$3x^{2}-432=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -432 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-432\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-432\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-432\right)}}{2\times 3}$$

Multiply -12 times -432.

$$x=\frac{0±\sqrt{5184}}{2\times 3}$$

Take the square root of 5184.

$$x=\frac{0±72}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±72}{6}$$

Now solve the equation x=\frac{0±72}{6} when ± is plus. Divide 72 by 6.

$$x=12$$

Now solve the equation x=\frac{0±72}{6} when ± is minus. Divide -72 by 6.

$$x=-12$$

The equation is now solved.

$$x=12 x=-12$$