Solve 10*6x=60x^2

$$10 \times 6x=60 { x }^{ 2 }$$

Solve for x

$$x=1$$

$$x=0$$

Steps Using Factoring

Multiply 10 and 6 to get 60.

$$60x=60x^{2}$$

Subtract 60x^{2} from both sides.

$$60x-60x^{2}=0$$

Factor out x.

$$x\left(60-60x\right)=0$$

To find equation solutions, solve x=0 and 60-60x=0.

$$x=0 x=1$$

Steps Using the Quadratic Formula

Multiply 10 and 6 to get 60.

$$60x=60x^{2}$$

Subtract 60x^{2} from both sides.

$$60x-60x^{2}=0$$

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.

$$-60x^{2}+60x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -60 for a, 60 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}}}{2\left(-60\right)}$$

Take the square root of 60^{2}.

$$x=\frac{-60±60}{2\left(-60\right)}$$

Multiply 2 times -60.

$$x=\frac{-60±60}{-120}$$

Now solve the equation x=\frac{-60±60}{-120} when ± is plus. Add -60 to 60.

$$x=\frac{0}{-120}$$

Divide 0 by -120.

$$x=0$$

Now solve the equation x=\frac{-60±60}{-120} when ± is minus. Subtract 60 from -60.

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

Divide -120 by -120.

$$x=1$$

The equation is now solved.

$$x=0 x=1$$

Steps for Completing the Square

Multiply 10 and 6 to get 60.

$$60x=60x^{2}$$

Subtract 60x^{2} from both sides.

$$60x-60x^{2}=0$$

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.

$$-60x^{2}+60x=0$$

Divide both sides by -60.

$$\frac{-60x^{2}+60x}{-60}=\frac{0}{-60}$$

Dividing by -60 undoes the multiplication by -60.

$$x^{2}+\frac{60}{-60}x=\frac{0}{-60}$$

Divide 60 by -60.

$$x^{2}-x=\frac{0}{-60}$$

Divide 0 by -60.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{1}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{1}{2} x-\frac{1}{2}=-\frac{1}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$