Solve 12x^2-160x+400=0

$$12 { x }^{ 2 } -160x+400 = 0$$

Solve for x

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

$$x=10$$

Steps Using the Quadratic Formula

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.

$$12x^{2}-160x+400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -160 for b, and 400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-160\right)±\sqrt{\left(-160\right)^{2}-4\times 12\times 400}}{2\times 12}$$

Square -160.

$$x=\frac{-\left(-160\right)±\sqrt{25600-4\times 12\times 400}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-160\right)±\sqrt{25600-48\times 400}}{2\times 12}$$

Multiply -48 times 400.

$$x=\frac{-\left(-160\right)±\sqrt{25600-19200}}{2\times 12}$$

Add 25600 to -19200.

$$x=\frac{-\left(-160\right)±\sqrt{6400}}{2\times 12}$$

Take the square root of 6400.

$$x=\frac{-\left(-160\right)±80}{2\times 12}$$

The opposite of -160 is 160.

$$x=\frac{160±80}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{160±80}{24}$$

Now solve the equation x=\frac{160±80}{24} when ± is plus. Add 160 to 80.

$$x=\frac{240}{24}$$

Divide 240 by 24.

$$x=10$$

Now solve the equation x=\frac{160±80}{24} when ± is minus. Subtract 80 from 160.

$$x=\frac{80}{24}$$

Reduce the fraction \frac{80}{24} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

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

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$12x^{2}-160x+400=0$$

Subtract 400 from both sides of the equation.

$$12x^{2}-160x+400-400=-400$$

Subtracting 400 from itself leaves 0.

$$12x^{2}-160x=-400$$

Divide both sides by 12.

$$\frac{12x^{2}-160x}{12}=-\frac{400}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{160}{12}\right)x=-\frac{400}{12}$$

Reduce the fraction \frac{-160}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{40}{3}x=-\frac{400}{12}$$

Reduce the fraction \frac{-400}{12} to lowest terms by extracting and canceling out 4.

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

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.

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