Solve 2x+3x^2-4

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

Factor

$$3\left(x-\frac{-\sqrt{13}-1}{3}\right)\left(x-\frac{\sqrt{13}-1}{3}\right)$$

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.

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

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.

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

Square 2.

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

Multiply -4 times 3.

$$x=\frac{-2±\sqrt{4-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$x=\frac{-2±\sqrt{4+48}}{2\times 3}$$

Add 4 to 48.

$$x=\frac{-2±\sqrt{52}}{2\times 3}$$

Take the square root of 52.

$$x=\frac{-2±2\sqrt{13}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-2±2\sqrt{13}}{6}$$

Now solve the equation x=\frac{-2±2\sqrt{13}}{6} when ± is plus. Add -2 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}-2}{6}$$

Divide -2+2\sqrt{13} by 6.

$$x=\frac{\sqrt{13}-1}{3}$$

Now solve the equation x=\frac{-2±2\sqrt{13}}{6} when ± is minus. Subtract 2\sqrt{13} from -2.

$$x=\frac{-2\sqrt{13}-2}{6}$$

Divide -2-2\sqrt{13} by 6.

$$x=\frac{-\sqrt{13}-1}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{13}}{3} for x_{1} and \frac{-1-\sqrt{13}}{3} for x_{2}.

$$3x^{2}+2x-4=3\left(x-\frac{\sqrt{13}-1}{3}\right)\left(x-\frac{-\sqrt{13}-1}{3}\right)$$

Evaluate

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