Solve 2x^2+10x-8

$$2 { x }^{ 2 } +10x-8$$

Factor

$$2\left(x-\frac{-\sqrt{41}-5}{2}\right)\left(x-\frac{\sqrt{41}-5}{2}\right)$$

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.

$$2x^{2}+10x-8=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8\left(-8\right)}}{2\times 2}$$

Multiply -8 times -8.

$$x=\frac{-10±\sqrt{100+64}}{2\times 2}$$

Add 100 to 64.

$$x=\frac{-10±\sqrt{164}}{2\times 2}$$

Take the square root of 164.

$$x=\frac{-10±2\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-10±2\sqrt{41}}{4}$$

Now solve the equation x=\frac{-10±2\sqrt{41}}{4} when ± is plus. Add -10 to 2\sqrt{41}.

$$x=\frac{2\sqrt{41}-10}{4}$$

Divide -10+2\sqrt{41} by 4.

$$x=\frac{\sqrt{41}-5}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{41}}{4} when ± is minus. Subtract 2\sqrt{41} from -10.

$$x=\frac{-2\sqrt{41}-10}{4}$$

Divide -10-2\sqrt{41} by 4.

$$x=\frac{-\sqrt{41}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{41}}{2} for x_{1} and \frac{-5-\sqrt{41}}{2} for x_{2}.

$$2x^{2}+10x-8=2\left(x-\frac{\sqrt{41}-5}{2}\right)\left(x-\frac{-\sqrt{41}-5}{2}\right)$$

Evaluate

$$2\left(x^{2}+5x-4\right)$$