Solve 40x^2+40=82x

$$40 x ^ { 2 } + 40 = 82 x$$

Solve for x

$$x=\frac{4}{5}=0.8$$

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using the Quadratic Formula

Subtract 82x from both sides.

$$40x^{2}+40-82x=0$$

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.

$$40x^{2}-82x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, -82 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-82\right)±\sqrt{\left(-82\right)^{2}-4\times 40\times 40}}{2\times 40}$$

Square -82.

$$x=\frac{-\left(-82\right)±\sqrt{6724-4\times 40\times 40}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-\left(-82\right)±\sqrt{6724-160\times 40}}{2\times 40}$$

Multiply -160 times 40.

$$x=\frac{-\left(-82\right)±\sqrt{6724-6400}}{2\times 40}$$

Add 6724 to -6400.

$$x=\frac{-\left(-82\right)±\sqrt{324}}{2\times 40}$$

Take the square root of 324.

$$x=\frac{-\left(-82\right)±18}{2\times 40}$$

The opposite of -82 is 82.

$$x=\frac{82±18}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{82±18}{80}$$

Now solve the equation x=\frac{82±18}{80} when ± is plus. Add 82 to 18.

$$x=\frac{100}{80}$$

Reduce the fraction \frac{100}{80} to lowest terms by extracting and canceling out 20.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{82±18}{80} when ± is minus. Subtract 18 from 82.

$$x=\frac{64}{80}$$

Reduce the fraction \frac{64}{80} to lowest terms by extracting and canceling out 16.

$$x=\frac{4}{5}$$

The equation is now solved.

$$x=\frac{5}{4} x=\frac{4}{5}$$

Steps for Completing the Square

Subtract 82x from both sides.

$$40x^{2}+40-82x=0$$

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

$$40x^{2}-82x=-40$$

Divide both sides by 40.

$$\frac{40x^{2}-82x}{40}=-\frac{40}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$x^{2}+\left(-\frac{82}{40}\right)x=-\frac{40}{40}$$

Reduce the fraction \frac{-82}{40} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{41}{20}x=-\frac{40}{40}$$

Divide -40 by 40.

$$x^{2}-\frac{41}{20}x=-1$$

Divide -\frac{41}{20}, the coefficient of the x term, by 2 to get -\frac{41}{40}. Then add the square of -\frac{41}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{41}{20}x+\left(-\frac{41}{40}\right)^{2}=-1+\left(-\frac{41}{40}\right)^{2}$$

Square -\frac{41}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{41}{20}x+\frac{1681}{1600}=-1+\frac{1681}{1600}$$

Add -1 to \frac{1681}{1600}.

$$x^{2}-\frac{41}{20}x+\frac{1681}{1600}=\frac{81}{1600}$$

Factor x^{2}-\frac{41}{20}x+\frac{1681}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{41}{40}\right)^{2}=\frac{81}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{40}\right)^{2}}=\sqrt{\frac{81}{1600}}$$

Simplify.

$$x-\frac{41}{40}=\frac{9}{40} x-\frac{41}{40}=-\frac{9}{40}$$

Add \frac{41}{40} to both sides of the equation.

$$x=\frac{5}{4} x=\frac{4}{5}$$