Solve 4x^2=36x-8

$$4 x ^ { 2 } = 36 x - 8$$

Solve for x

$$x = \frac{\sqrt{73} + 9}{2} \approx 8.772001873$$

$$x=\frac{9-\sqrt{73}}{2}\approx 0.227998127$$

Steps Using the Quadratic Formula

Subtract 36x from both sides.

$$4x^{2}-36x=-8$$

Add 8 to both sides.

$$4x^{2}-36x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -36 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 4\times 8}}{2\times 4}$$

Square -36.

$$x=\frac{-\left(-36\right)±\sqrt{1296-4\times 4\times 8}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-36\right)±\sqrt{1296-16\times 8}}{2\times 4}$$

Multiply -16 times 8.

$$x=\frac{-\left(-36\right)±\sqrt{1296-128}}{2\times 4}$$

Add 1296 to -128.

$$x=\frac{-\left(-36\right)±\sqrt{1168}}{2\times 4}$$

Take the square root of 1168.

$$x=\frac{-\left(-36\right)±4\sqrt{73}}{2\times 4}$$

The opposite of -36 is 36.

$$x=\frac{36±4\sqrt{73}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{36±4\sqrt{73}}{8}$$

Now solve the equation x=\frac{36±4\sqrt{73}}{8} when ± is plus. Add 36 to 4\sqrt{73}.

$$x=\frac{4\sqrt{73}+36}{8}$$

Divide 36+4\sqrt{73} by 8.

$$x=\frac{\sqrt{73}+9}{2}$$

Now solve the equation x=\frac{36±4\sqrt{73}}{8} when ± is minus. Subtract 4\sqrt{73} from 36.

$$x=\frac{36-4\sqrt{73}}{8}$$

Divide 36-4\sqrt{73} by 8.

$$x=\frac{9-\sqrt{73}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{73}+9}{2} x=\frac{9-\sqrt{73}}{2}$$

Steps for Completing the Square

Subtract 36x from both sides.

$$4x^{2}-36x=-8$$

Divide both sides by 4.

$$\frac{4x^{2}-36x}{4}=-\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{36}{4}\right)x=-\frac{8}{4}$$

Divide -36 by 4.

$$x^{2}-9x=-\frac{8}{4}$$

Divide -8 by 4.

$$x^{2}-9x=-2$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-2+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-2+\frac{81}{4}$$

Add -2 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{73}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{73}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{73}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{\sqrt{73}}{2} x-\frac{9}{2}=-\frac{\sqrt{73}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=\frac{\sqrt{73}+9}{2} x=\frac{9-\sqrt{73}}{2}$$