Solve x=4x^2+2x-17

$$x=4 { x }^{ 2 } +2x-17$$

Solve for x

$$x = \frac{\sqrt{273} - 1}{8} \approx 1.940338955$$

$$x=\frac{-\sqrt{273}-1}{8}\approx -2.190338955$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$x-4x^{2}=2x-17$$

Subtract 2x from both sides.

$$x-4x^{2}-2x=-17$$

Combine x and -2x to get -x.

$$-x-4x^{2}=-17$$

Add 17 to both sides.

$$-x-4x^{2}+17=0$$

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.

$$-4x^{2}-x+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -1 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-4\right)\times 17}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-1\right)±\sqrt{1+16\times 17}}{2\left(-4\right)}$$

Multiply 16 times 17.

$$x=\frac{-\left(-1\right)±\sqrt{1+272}}{2\left(-4\right)}$$

Add 1 to 272.

$$x=\frac{-\left(-1\right)±\sqrt{273}}{2\left(-4\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{273}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{1±\sqrt{273}}{-8}$$

Now solve the equation x=\frac{1±\sqrt{273}}{-8} when ± is plus. Add 1 to \sqrt{273}.

$$x=\frac{\sqrt{273}+1}{-8}$$

Divide 1+\sqrt{273} by -8.

$$x=\frac{-\sqrt{273}-1}{8}$$

Now solve the equation x=\frac{1±\sqrt{273}}{-8} when ± is minus. Subtract \sqrt{273} from 1.

$$x=\frac{1-\sqrt{273}}{-8}$$

Divide 1-\sqrt{273} by -8.

$$x=\frac{\sqrt{273}-1}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{273}-1}{8} x=\frac{\sqrt{273}-1}{8}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$x-4x^{2}=2x-17$$

Subtract 2x from both sides.

$$x-4x^{2}-2x=-17$$

Combine x and -2x to get -x.

$$-x-4x^{2}=-17$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4x^{2}-x=-17$$

Divide both sides by -4.

$$\frac{-4x^{2}-x}{-4}=-\frac{17}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{1}{-4}\right)x=-\frac{17}{-4}$$

Divide -1 by -4.

$$x^{2}+\frac{1}{4}x=-\frac{17}{-4}$$

Divide -17 by -4.

$$x^{2}+\frac{1}{4}x=\frac{17}{4}$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{17}{4}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{17}{4}+\frac{1}{64}$$

Add \frac{17}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{273}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{273}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{273}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{\sqrt{273}}{8} x+\frac{1}{8}=-\frac{\sqrt{273}}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=\frac{\sqrt{273}-1}{8} x=\frac{-\sqrt{273}-1}{8}$$