Solve 4x^2+5x-8

$$4 { x }^{ 2 } +5x-8$$

Factor

$$4\left(x-\frac{-3\sqrt{17}-5}{8}\right)\left(x-\frac{3\sqrt{17}-5}{8}\right)$$

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.

$$4x^{2}+5x-8=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-5±\sqrt{25-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$x=\frac{-5±\sqrt{25+128}}{2\times 4}$$

Add 25 to 128.

$$x=\frac{-5±\sqrt{153}}{2\times 4}$$

Take the square root of 153.

$$x=\frac{-5±3\sqrt{17}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-5±3\sqrt{17}}{8}$$

Now solve the equation x=\frac{-5±3\sqrt{17}}{8} when ± is plus. Add -5 to 3\sqrt{17}.

$$x=\frac{3\sqrt{17}-5}{8}$$

Now solve the equation x=\frac{-5±3\sqrt{17}}{8} when ± is minus. Subtract 3\sqrt{17} from -5.

$$x=\frac{-3\sqrt{17}-5}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+3\sqrt{17}}{8} for x_{1} and \frac{-5-3\sqrt{17}}{8} for x_{2}.

$$4x^{2}+5x-8=4\left(x-\frac{3\sqrt{17}-5}{8}\right)\left(x-\frac{-3\sqrt{17}-5}{8}\right)$$

Evaluate

$$4x^{2}+5x-8$$