Solve 4x^2=40x+8

$$4 x ^ { 2 } = 40 x + 8$$

Solve for x

$$x=3\sqrt{3}+5\approx 10.196152423$$

$$x=5-3\sqrt{3}\approx -0.196152423$$

Steps Using the Quadratic Formula

Subtract 40x from both sides.

$$4x^{2}-40x=8$$

Subtract 8 from both sides.

$$4x^{2}-40x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -40 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-40\right)±\sqrt{1600-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$x=\frac{-\left(-40\right)±\sqrt{1600+128}}{2\times 4}$$

Add 1600 to 128.

$$x=\frac{-\left(-40\right)±\sqrt{1728}}{2\times 4}$$

Take the square root of 1728.

$$x=\frac{-\left(-40\right)±24\sqrt{3}}{2\times 4}$$

The opposite of -40 is 40.

$$x=\frac{40±24\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{40±24\sqrt{3}}{8}$$

Now solve the equation x=\frac{40±24\sqrt{3}}{8} when ± is plus. Add 40 to 24\sqrt{3}.

$$x=\frac{24\sqrt{3}+40}{8}$$

Divide 40+24\sqrt{3} by 8.

$$x=3\sqrt{3}+5$$

Now solve the equation x=\frac{40±24\sqrt{3}}{8} when ± is minus. Subtract 24\sqrt{3} from 40.

$$x=\frac{40-24\sqrt{3}}{8}$$

Divide 40-24\sqrt{3} by 8.

$$x=5-3\sqrt{3}$$

The equation is now solved.

$$x=3\sqrt{3}+5 x=5-3\sqrt{3}$$

Steps for Completing the Square

Subtract 40x from both sides.

$$4x^{2}-40x=8$$

Divide both sides by 4.

$$\frac{4x^{2}-40x}{4}=\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{40}{4}\right)x=\frac{8}{4}$$

Divide -40 by 4.

$$x^{2}-10x=\frac{8}{4}$$

Divide 8 by 4.

$$x^{2}-10x=2$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -5.

$$x^{2}-10x+25=2+25$$

Add 2 to 25.

$$x^{2}-10x+25=27$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=27$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{27}$$

Simplify.

$$x-5=3\sqrt{3} x-5=-3\sqrt{3}$$

Add 5 to both sides of the equation.

$$x=3\sqrt{3}+5 x=5-3\sqrt{3}$$