Solve 300=-x^2+25x

$$300 = - x ^ { 2 } + 25 x$$

Solve for x (complex solution)

$$x=\frac{25+5\sqrt{23}i}{2}\approx 12.5+11.989578808i$$

$$x=\frac{-5\sqrt{23}i+25}{2}\approx 12.5-11.989578808i$$

Steps Using the Quadratic Formula

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

$$-x^{2}+25x=300$$

Subtract 300 from both sides.

$$-x^{2}+25x-300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 25 for b, and -300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-1\right)\left(-300\right)}}{2\left(-1\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-1\right)\left(-300\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-25±\sqrt{625+4\left(-300\right)}}{2\left(-1\right)}$$

Multiply 4 times -300.

$$x=\frac{-25±\sqrt{625-1200}}{2\left(-1\right)}$$

Add 625 to -1200.

$$x=\frac{-25±\sqrt{-575}}{2\left(-1\right)}$$

Take the square root of -575.

$$x=\frac{-25±5\sqrt{23}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-25±5\sqrt{23}i}{-2}$$

Now solve the equation x=\frac{-25±5\sqrt{23}i}{-2} when ± is plus. Add -25 to 5i\sqrt{23}.

$$x=\frac{-25+5\sqrt{23}i}{-2}$$

Divide -25+5i\sqrt{23} by -2.

$$x=\frac{-5\sqrt{23}i+25}{2}$$

Now solve the equation x=\frac{-25±5\sqrt{23}i}{-2} when ± is minus. Subtract 5i\sqrt{23} from -25.

$$x=\frac{-5\sqrt{23}i-25}{-2}$$

Divide -25-5i\sqrt{23} by -2.

$$x=\frac{25+5\sqrt{23}i}{2}$$

The equation is now solved.

$$x=\frac{-5\sqrt{23}i+25}{2} x=\frac{25+5\sqrt{23}i}{2}$$

Steps for Completing the Square

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

$$-x^{2}+25x=300$$

Divide both sides by -1.

$$\frac{-x^{2}+25x}{-1}=\frac{300}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{25}{-1}x=\frac{300}{-1}$$

Divide 25 by -1.

$$x^{2}-25x=\frac{300}{-1}$$

Divide 300 by -1.

$$x^{2}-25x=-300$$

Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-25x+\left(-\frac{25}{2}\right)^{2}=-300+\left(-\frac{25}{2}\right)^{2}$$

Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-25x+\frac{625}{4}=-300+\frac{625}{4}$$

Add -300 to \frac{625}{4}.

$$x^{2}-25x+\frac{625}{4}=-\frac{575}{4}$$

Factor x^{2}-25x+\frac{625}{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}.

$$\left(x-\frac{25}{2}\right)^{2}=-\frac{575}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{2}\right)^{2}}=\sqrt{-\frac{575}{4}}$$

Simplify.

$$x-\frac{25}{2}=\frac{5\sqrt{23}i}{2} x-\frac{25}{2}=-\frac{5\sqrt{23}i}{2}$$

Add \frac{25}{2} to both sides of the equation.

$$x=\frac{25+5\sqrt{23}i}{2} x=\frac{-5\sqrt{23}i+25}{2}$$