Solve 200+6x^2=80x

$$200 + 6 x ^ { 2 } = 80 x$$

Solve for x

$$x = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

$$x=10$$

Steps Using the Quadratic Formula

Subtract 80x from both sides.

$$200+6x^{2}-80x=0$$

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.

$$6x^{2}-80x+200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -80 for b, and 200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 6\times 200}}{2\times 6}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 6\times 200}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-80\right)±\sqrt{6400-24\times 200}}{2\times 6}$$

Multiply -24 times 200.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4800}}{2\times 6}$$

Add 6400 to -4800.

$$x=\frac{-\left(-80\right)±\sqrt{1600}}{2\times 6}$$

Take the square root of 1600.

$$x=\frac{-\left(-80\right)±40}{2\times 6}$$

The opposite of -80 is 80.

$$x=\frac{80±40}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{80±40}{12}$$

Now solve the equation x=\frac{80±40}{12} when ± is plus. Add 80 to 40.

$$x=\frac{120}{12}$$

Divide 120 by 12.

$$x=10$$

Now solve the equation x=\frac{80±40}{12} when ± is minus. Subtract 40 from 80.

$$x=\frac{40}{12}$$

Reduce the fraction \frac{40}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{10}{3}$$

The equation is now solved.

$$x=10 x=\frac{10}{3}$$

Steps for Completing the Square

Subtract 80x from both sides.

$$200+6x^{2}-80x=0$$

Subtract 200 from both sides. Anything subtracted from zero gives its negation.

$$6x^{2}-80x=-200$$

Divide both sides by 6.

$$\frac{6x^{2}-80x}{6}=-\frac{200}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{80}{6}\right)x=-\frac{200}{6}$$

Reduce the fraction \frac{-80}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{40}{3}x=-\frac{200}{6}$$

Reduce the fraction \frac{-200}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{40}{3}x=-\frac{100}{3}$$

Divide -\frac{40}{3}, the coefficient of the x term, by 2 to get -\frac{20}{3}. Then add the square of -\frac{20}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{40}{3}x+\left(-\frac{20}{3}\right)^{2}=-\frac{100}{3}+\left(-\frac{20}{3}\right)^{2}$$

Square -\frac{20}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{40}{3}x+\frac{400}{9}=-\frac{100}{3}+\frac{400}{9}$$

Add -\frac{100}{3} to \frac{400}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{40}{3}x+\frac{400}{9}=\frac{100}{9}$$

Factor x^{2}-\frac{40}{3}x+\frac{400}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{20}{3}\right)^{2}=\frac{100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{20}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Simplify.

$$x-\frac{20}{3}=\frac{10}{3} x-\frac{20}{3}=-\frac{10}{3}$$

Add \frac{20}{3} to both sides of the equation.

$$x=10 x=\frac{10}{3}$$