Solve 12x^2+9x-6

$$12 { x }^{ 2 } +9x-6$$

Factor

$$12\left(x-\frac{-\sqrt{41}-3}{8}\right)\left(x-\frac{\sqrt{41}-3}{8}\right)$$

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.

$$12x^{2}+9x-6=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\times 12\left(-6\right)}}{2\times 12}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 12\left(-6\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-9±\sqrt{81-48\left(-6\right)}}{2\times 12}$$

Multiply -48 times -6.

$$x=\frac{-9±\sqrt{81+288}}{2\times 12}$$

Add 81 to 288.

$$x=\frac{-9±\sqrt{369}}{2\times 12}$$

Take the square root of 369.

$$x=\frac{-9±3\sqrt{41}}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-9±3\sqrt{41}}{24}$$

Now solve the equation x=\frac{-9±3\sqrt{41}}{24} when ± is plus. Add -9 to 3\sqrt{41}.

$$x=\frac{3\sqrt{41}-9}{24}$$

Divide -9+3\sqrt{41} by 24.

$$x=\frac{\sqrt{41}-3}{8}$$

Now solve the equation x=\frac{-9±3\sqrt{41}}{24} when ± is minus. Subtract 3\sqrt{41} from -9.

$$x=\frac{-3\sqrt{41}-9}{24}$$

Divide -9-3\sqrt{41} by 24.

$$x=\frac{-\sqrt{41}-3}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{41}}{8} for x_{1} and \frac{-3-\sqrt{41}}{8} for x_{2}.

$$12x^{2}+9x-6=12\left(x-\frac{\sqrt{41}-3}{8}\right)\left(x-\frac{-\sqrt{41}-3}{8}\right)$$

Evaluate

$$12x^{2}+9x-6$$