Solve 40+0.085x^2=5x

$$40 + 0.085 x ^ { 2 } = 5 x$$

Solve for x

$$x = \frac{20 \sqrt{285} + 500}{17} \approx 49.272874137$$

$$x = \frac{500 - 20 \sqrt{285}}{17} \approx 9.550655275$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

$$40+0.085x^{2}-5x=0$$

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.

$$0.085x^{2}-5x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.085 for a, -5 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 0.085\times 40}}{2\times 0.085}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 0.085\times 40}}{2\times 0.085}$$

Multiply -4 times 0.085.

$$x=\frac{-\left(-5\right)±\sqrt{25-0.34\times 40}}{2\times 0.085}$$

Multiply -0.34 times 40.

$$x=\frac{-\left(-5\right)±\sqrt{25-13.6}}{2\times 0.085}$$

Add 25 to -13.6.

$$x=\frac{-\left(-5\right)±\sqrt{11.4}}{2\times 0.085}$$

Take the square root of 11.4.

$$x=\frac{-\left(-5\right)±\frac{\sqrt{285}}{5}}{2\times 0.085}$$

The opposite of -5 is 5.

$$x=\frac{5±\frac{\sqrt{285}}{5}}{2\times 0.085}$$

Multiply 2 times 0.085.

$$x=\frac{5±\frac{\sqrt{285}}{5}}{0.17}$$

Now solve the equation x=\frac{5±\frac{\sqrt{285}}{5}}{0.17} when ± is plus. Add 5 to \frac{\sqrt{285}}{5}.

$$x=\frac{\frac{\sqrt{285}}{5}+5}{0.17}$$

Divide 5+\frac{\sqrt{285}}{5} by 0.17 by multiplying 5+\frac{\sqrt{285}}{5} by the reciprocal of 0.17.

$$x=\frac{20\sqrt{285}+500}{17}$$

Now solve the equation x=\frac{5±\frac{\sqrt{285}}{5}}{0.17} when ± is minus. Subtract \frac{\sqrt{285}}{5} from 5.

$$x=\frac{-\frac{\sqrt{285}}{5}+5}{0.17}$$

Divide 5-\frac{\sqrt{285}}{5} by 0.17 by multiplying 5-\frac{\sqrt{285}}{5} by the reciprocal of 0.17.

$$x=\frac{500-20\sqrt{285}}{17}$$

The equation is now solved.

$$x=\frac{20\sqrt{285}+500}{17} x=\frac{500-20\sqrt{285}}{17}$$

Steps for Completing the Square

Subtract 5x from both sides.

$$40+0.085x^{2}-5x=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$0.085x^{2}-5x=-40$$

Divide both sides of the equation by 0.085, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{0.085x^{2}-5x}{0.085}=-\frac{40}{0.085}$$

Dividing by 0.085 undoes the multiplication by 0.085.

$$x^{2}+\left(-\frac{5}{0.085}\right)x=-\frac{40}{0.085}$$

Divide -5 by 0.085 by multiplying -5 by the reciprocal of 0.085.

$$x^{2}-\frac{1000}{17}x=-\frac{40}{0.085}$$

Divide -40 by 0.085 by multiplying -40 by the reciprocal of 0.085.

$$x^{2}-\frac{1000}{17}x=-\frac{8000}{17}$$

Divide -\frac{1000}{17}, the coefficient of the x term, by 2 to get -\frac{500}{17}. Then add the square of -\frac{500}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1000}{17}x+\left(-\frac{500}{17}\right)^{2}=-\frac{8000}{17}+\left(-\frac{500}{17}\right)^{2}$$

Square -\frac{500}{17} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1000}{17}x+\frac{250000}{289}=-\frac{8000}{17}+\frac{250000}{289}$$

Add -\frac{8000}{17} to \frac{250000}{289} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1000}{17}x+\frac{250000}{289}=\frac{114000}{289}$$

Factor x^{2}-\frac{1000}{17}x+\frac{250000}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{500}{17}\right)^{2}=\frac{114000}{289}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{500}{17}\right)^{2}}=\sqrt{\frac{114000}{289}}$$

Simplify.

$$x-\frac{500}{17}=\frac{20\sqrt{285}}{17} x-\frac{500}{17}=-\frac{20\sqrt{285}}{17}$$

Add \frac{500}{17} to both sides of the equation.

$$x=\frac{20\sqrt{285}+500}{17} x=\frac{500-20\sqrt{285}}{17}$$