Solve 3x+6x^2-4

$$3x+6 { x }^{ 2 } -4$$

Factor

$$6\left(x-\left(-\frac{\sqrt{105}}{12}-\frac{1}{4}\right)\right)\left(x-\left(\frac{\sqrt{105}}{12}-\frac{1}{4}\right)\right)$$

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.

$$6x^{2}+3x-4=0$$

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.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 6\left(-4\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-3±\sqrt{9-24\left(-4\right)}}{2\times 6}$$

Multiply -24 times -4.

$$x=\frac{-3±\sqrt{9+96}}{2\times 6}$$

Add 9 to 96.

$$x=\frac{-3±\sqrt{105}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-3±\sqrt{105}}{12}$$

Now solve the equation x=\frac{-3±\sqrt{105}}{12} when ± is plus. Add -3 to \sqrt{105}.

$$x=\frac{\sqrt{105}-3}{12}$$

Divide -3+\sqrt{105} by 12.

$$x=\frac{\sqrt{105}}{12}-\frac{1}{4}$$

Now solve the equation x=\frac{-3±\sqrt{105}}{12} when ± is minus. Subtract \sqrt{105} from -3.

$$x=\frac{-\sqrt{105}-3}{12}$$

Divide -3-\sqrt{105} by 12.

$$x=-\frac{\sqrt{105}}{12}-\frac{1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{4}+\frac{\sqrt{105}}{12} for x_{1} and -\frac{1}{4}-\frac{\sqrt{105}}{12} for x_{2}.

$$6x^{2}+3x-4=6\left(x-\left(\frac{\sqrt{105}}{12}-\frac{1}{4}\right)\right)\left(x-\left(-\frac{\sqrt{105}}{12}-\frac{1}{4}\right)\right)$$

Evaluate

$$6x^{2}+3x-4$$