Solve x-4+4x^2

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

Factor

$$4\left(x-\frac{-\sqrt{65}-1}{8}\right)\left(x-\frac{\sqrt{65}-1}{8}\right)$$

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.

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

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-4\right)}}{2\times 4}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-4\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-1±\sqrt{1-16\left(-4\right)}}{2\times 4}$$

Multiply -16 times -4.

$$x=\frac{-1±\sqrt{1+64}}{2\times 4}$$

Add 1 to 64.

$$x=\frac{-1±\sqrt{65}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-1±\sqrt{65}}{8}$$

Now solve the equation x=\frac{-1±\sqrt{65}}{8} when ± is plus. Add -1 to \sqrt{65}.

$$x=\frac{\sqrt{65}-1}{8}$$

Now solve the equation x=\frac{-1±\sqrt{65}}{8} when ± is minus. Subtract \sqrt{65} from -1.

$$x=\frac{-\sqrt{65}-1}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{65}}{8} for x_{1} and \frac{-1-\sqrt{65}}{8} for x_{2}.

$$4x^{2}+x-4=4\left(x-\frac{\sqrt{65}-1}{8}\right)\left(x-\frac{-\sqrt{65}-1}{8}\right)$$

Evaluate

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