Solve 5x^2=40x+5

$$5 x ^ { 2 } = 40 x + 5$$

Solve for x

$$x=\sqrt{17}+4\approx 8.123105626$$

$$x=4-\sqrt{17}\approx -0.123105626$$

Steps Using the Quadratic Formula

Subtract 40x from both sides.

$$5x^{2}-40x=5$$

Subtract 5 from both sides.

$$5x^{2}-40x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -40 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\left(-5\right)}}{2\times 5}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\left(-5\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-40\right)±\sqrt{1600-20\left(-5\right)}}{2\times 5}$$

Multiply -20 times -5.

$$x=\frac{-\left(-40\right)±\sqrt{1600+100}}{2\times 5}$$

Add 1600 to 100.

$$x=\frac{-\left(-40\right)±\sqrt{1700}}{2\times 5}$$

Take the square root of 1700.

$$x=\frac{-\left(-40\right)±10\sqrt{17}}{2\times 5}$$

The opposite of -40 is 40.

$$x=\frac{40±10\sqrt{17}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{40±10\sqrt{17}}{10}$$

Now solve the equation x=\frac{40±10\sqrt{17}}{10} when ± is plus. Add 40 to 10\sqrt{17}.

$$x=\frac{10\sqrt{17}+40}{10}$$

Divide 40+10\sqrt{17} by 10.

$$x=\sqrt{17}+4$$

Now solve the equation x=\frac{40±10\sqrt{17}}{10} when ± is minus. Subtract 10\sqrt{17} from 40.

$$x=\frac{40-10\sqrt{17}}{10}$$

Divide 40-10\sqrt{17} by 10.

$$x=4-\sqrt{17}$$

The equation is now solved.

$$x=\sqrt{17}+4 x=4-\sqrt{17}$$

Steps for Completing the Square

Subtract 40x from both sides.

$$5x^{2}-40x=5$$

Divide both sides by 5.

$$\frac{5x^{2}-40x}{5}=\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{40}{5}\right)x=\frac{5}{5}$$

Divide -40 by 5.

$$x^{2}-8x=\frac{5}{5}$$

Divide 5 by 5.

$$x^{2}-8x=1$$

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

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

Square -4.

$$x^{2}-8x+16=1+16$$

Add 1 to 16.

$$x^{2}-8x+16=17$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{17}$$

Simplify.

$$x-4=\sqrt{17} x-4=-\sqrt{17}$$

Add 4 to both sides of the equation.

$$x=\sqrt{17}+4 x=4-\sqrt{17}$$