Solve 2x^2+2x-3

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

Factor

$$2\left(x-\frac{-\sqrt{7}-1}{2}\right)\left(x-\frac{\sqrt{7}-1}{2}\right)$$

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.

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

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.

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

Square 2.

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

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-2±\sqrt{4+24}}{2\times 2}$$

Add 4 to 24.

$$x=\frac{-2±\sqrt{28}}{2\times 2}$$

Take the square root of 28.

$$x=\frac{-2±2\sqrt{7}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±2\sqrt{7}}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{7}}{4} when ± is plus. Add -2 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-2}{4}$$

Divide -2+2\sqrt{7} by 4.

$$x=\frac{\sqrt{7}-1}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{7}}{4} when ± is minus. Subtract 2\sqrt{7} from -2.

$$x=\frac{-2\sqrt{7}-2}{4}$$

Divide -2-2\sqrt{7} by 4.

$$x=\frac{-\sqrt{7}-1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{7}}{2} for x_{1} and \frac{-1-\sqrt{7}}{2} for x_{2}.

$$2x^{2}+2x-3=2\left(x-\frac{\sqrt{7}-1}{2}\right)\left(x-\frac{-\sqrt{7}-1}{2}\right)$$

Evaluate

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