Solve 0=-3/2x^2+60x

$$0=- \frac{ 3 }{ 2 } { x }^{ 2 } +60x$$

Solve for x

$$x=40$$

$$x=0$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$-\frac{3}{2}x^{2}+60x=0$$

Factor out x.

$$x\left(-\frac{3}{2}x+60\right)=0$$

To find equation solutions, solve x=0 and -\frac{3x}{2}+60=0.

$$x=0 x=40$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-\frac{3}{2}x^{2}+60x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{2} for a, 60 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}}}{2\left(-\frac{3}{2}\right)}$$

Take the square root of 60^{2}.

$$x=\frac{-60±60}{2\left(-\frac{3}{2}\right)}$$

Multiply 2 times -\frac{3}{2}.

$$x=\frac{-60±60}{-3}$$

Now solve the equation x=\frac{-60±60}{-3} when ± is plus. Add -60 to 60.

$$x=\frac{0}{-3}$$

Divide 0 by -3.

$$x=0$$

Now solve the equation x=\frac{-60±60}{-3} when ± is minus. Subtract 60 from -60.

$$x=-\frac{120}{-3}$$

Divide -120 by -3.

$$x=40$$

The equation is now solved.

$$x=0 x=40$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-\frac{3}{2}x^{2}+60x=0$$

Divide both sides of the equation by -\frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{2}x^{2}+60x}{-\frac{3}{2}}=\frac{0}{-\frac{3}{2}}$$

Dividing by -\frac{3}{2} undoes the multiplication by -\frac{3}{2}.

$$x^{2}+\frac{60}{-\frac{3}{2}}x=\frac{0}{-\frac{3}{2}}$$

Divide 60 by -\frac{3}{2} by multiplying 60 by the reciprocal of -\frac{3}{2}.

$$x^{2}-40x=\frac{0}{-\frac{3}{2}}$$

Divide 0 by -\frac{3}{2} by multiplying 0 by the reciprocal of -\frac{3}{2}.

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

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

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

Square -20.

$$x^{2}-40x+400=400$$

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

$$\left(x-20\right)^{2}=400$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-20\right)^{2}}=\sqrt{400}$$

Simplify.

$$x-20=20 x-20=-20$$

Add 20 to both sides of the equation.

$$x=40 x=0$$