Solve 0=-0.2x^2+3x

$$0 = - 0.2 x ^ { 2 } + 3 x$$

Solve for x

$$x=15$$

$$x=0$$

Steps Using Factoring

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

$$-0.2x^{2}+3x=0$$

Factor out x.

$$x\left(-0.2x+3\right)=0$$

To find equation solutions, solve x=0 and -\frac{x}{5}+3=0.

$$x=0 x=15$$

Steps Using the Quadratic Formula

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

$$-0.2x^{2}+3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.2 for a, 3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}}}{2\left(-0.2\right)}$$

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\left(-0.2\right)}$$

Multiply 2 times -0.2.

$$x=\frac{-3±3}{-0.4}$$

Now solve the equation x=\frac{-3±3}{-0.4} when ± is plus. Add -3 to 3.

$$x=\frac{0}{-0.4}$$

Divide 0 by -0.4 by multiplying 0 by the reciprocal of -0.4.

$$x=0$$

Now solve the equation x=\frac{-3±3}{-0.4} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{-0.4}$$

Divide -6 by -0.4 by multiplying -6 by the reciprocal of -0.4.

$$x=15$$

The equation is now solved.

$$x=0 x=15$$

Steps for Completing the Square

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

$$-0.2x^{2}+3x=0$$

Multiply both sides by -5.

$$\frac{-0.2x^{2}+3x}{-0.2}=\frac{0}{-0.2}$$

Dividing by -0.2 undoes the multiplication by -0.2.

$$x^{2}+\frac{3}{-0.2}x=\frac{0}{-0.2}$$

Divide 3 by -0.2 by multiplying 3 by the reciprocal of -0.2.

$$x^{2}-15x=\frac{0}{-0.2}$$

Divide 0 by -0.2 by multiplying 0 by the reciprocal of -0.2.

$$x^{2}-15x=0$$

Divide -15, the coefficient of the x term, by 2 to get -\frac{15}{2}. Then add the square of -\frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-15x+\left(-\frac{15}{2}\right)^{2}=\left(-\frac{15}{2}\right)^{2}$$

Square -\frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-15x+\frac{225}{4}=\frac{225}{4}$$

Factor x^{2}-15x+\frac{225}{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}.

$$\left(x-\frac{15}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x-\frac{15}{2}=\frac{15}{2} x-\frac{15}{2}=-\frac{15}{2}$$

Add \frac{15}{2} to both sides of the equation.

$$x=15 x=0$$