Solve 3x^2+18x+18

$$3 { x }^{ 2 } +18x+18$$

Factor

$$3\left(x-\left(-\sqrt{3}-3\right)\right)\left(x-\left(\sqrt{3}-3\right)\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$3x^{2}+18x+18=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-18±\sqrt{18^{2}-4\times 3\times 18}}{2\times 3}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 3\times 18}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-18±\sqrt{324-12\times 18}}{2\times 3}$$

Multiply -12 times 18.

$$x=\frac{-18±\sqrt{324-216}}{2\times 3}$$

Add 324 to -216.

$$x=\frac{-18±\sqrt{108}}{2\times 3}$$

Take the square root of 108.

$$x=\frac{-18±6\sqrt{3}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-18±6\sqrt{3}}{6}$$

Now solve the equation x=\frac{-18±6\sqrt{3}}{6} when ± is plus. Add -18 to 6\sqrt{3}.

$$x=\frac{6\sqrt{3}-18}{6}$$

Divide -18+6\sqrt{3} by 6.

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

Now solve the equation x=\frac{-18±6\sqrt{3}}{6} when ± is minus. Subtract 6\sqrt{3} from -18.

$$x=\frac{-6\sqrt{3}-18}{6}$$

Divide -18-6\sqrt{3} by 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\sqrt{3} for x_{1} and -3-\sqrt{3} for x_{2}.

$$3x^{2}+18x+18=3\left(x-\left(\sqrt{3}-3\right)\right)\left(x-\left(-\sqrt{3}-3\right)\right)$$

Evaluate

$$3\left(x^{2}+6x+6\right)$$