Solve -0.3x^2+6x=25

$$-0.3 { x }^{ 2 } +6x=25$$

Solve for x

$$x=\frac{5\sqrt{6}}{3}+10\approx 14.082482905$$

$$x=-\frac{5\sqrt{6}}{3}+10\approx 5.917517095$$

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.

$$-0.3x^{2}+6x=25$$

Subtract 25 from both sides of the equation.

$$-0.3x^{2}+6x-25=25-25$$

Subtracting 25 from itself leaves 0.

$$-0.3x^{2}+6x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.3 for a, 6 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-0.3\right)\left(-25\right)}}{2\left(-0.3\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-0.3\right)\left(-25\right)}}{2\left(-0.3\right)}$$

Multiply -4 times -0.3.

$$x=\frac{-6±\sqrt{36+1.2\left(-25\right)}}{2\left(-0.3\right)}$$

Multiply 1.2 times -25.

$$x=\frac{-6±\sqrt{36-30}}{2\left(-0.3\right)}$$

Add 36 to -30.

$$x=\frac{-6±\sqrt{6}}{2\left(-0.3\right)}$$

Multiply 2 times -0.3.

$$x=\frac{-6±\sqrt{6}}{-0.6}$$

Now solve the equation x=\frac{-6±\sqrt{6}}{-0.6} when ± is plus. Add -6 to \sqrt{6}.

$$x=\frac{\sqrt{6}-6}{-0.6}$$

Divide -6+\sqrt{6} by -0.6 by multiplying -6+\sqrt{6} by the reciprocal of -0.6.

$$x=-\frac{5\sqrt{6}}{3}+10$$

Now solve the equation x=\frac{-6±\sqrt{6}}{-0.6} when ± is minus. Subtract \sqrt{6} from -6.

$$x=\frac{-\sqrt{6}-6}{-0.6}$$

Divide -6-\sqrt{6} by -0.6 by multiplying -6-\sqrt{6} by the reciprocal of -0.6.

$$x=\frac{5\sqrt{6}}{3}+10$$

The equation is now solved.

$$x=-\frac{5\sqrt{6}}{3}+10 x=\frac{5\sqrt{6}}{3}+10$$

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.

$$-0.3x^{2}+6x=25$$

Divide both sides of the equation by -0.3, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-0.3x^{2}+6x}{-0.3}=\frac{25}{-0.3}$$

Dividing by -0.3 undoes the multiplication by -0.3.

$$x^{2}+\frac{6}{-0.3}x=\frac{25}{-0.3}$$

Divide 6 by -0.3 by multiplying 6 by the reciprocal of -0.3.

$$x^{2}-20x=\frac{25}{-0.3}$$

Divide 25 by -0.3 by multiplying 25 by the reciprocal of -0.3.

$$x^{2}-20x=-\frac{250}{3}$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-20x+\left(-10\right)^{2}=-\frac{250}{3}+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=-\frac{250}{3}+100$$

Add -\frac{250}{3} to 100.

$$x^{2}-20x+100=\frac{50}{3}$$

Factor x^{2}-20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-10\right)^{2}=\frac{50}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{\frac{50}{3}}$$

Simplify.

$$x-10=\frac{5\sqrt{6}}{3} x-10=-\frac{5\sqrt{6}}{3}$$

Add 10 to both sides of the equation.

$$x=\frac{5\sqrt{6}}{3}+10 x=-\frac{5\sqrt{6}}{3}+10$$