Solve 625=15x+-10x^2/2

$$625=15x+ \frac{ -10 { x }^{ 2 } }{ 2 }$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{491}i}{2}\approx 1.5+11.079259903i$$

$$x=\frac{-\sqrt{491}i+3}{2}\approx 1.5-11.079259903i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$1250=30x-10x^{2}$$

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

$$30x-10x^{2}=1250$$

Subtract 1250 from both sides.

$$30x-10x^{2}-1250=0$$

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.

$$-10x^{2}+30x-1250=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 30 for b, and -1250 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-10\right)\left(-1250\right)}}{2\left(-10\right)}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-10\right)\left(-1250\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-30±\sqrt{900+40\left(-1250\right)}}{2\left(-10\right)}$$

Multiply 40 times -1250.

$$x=\frac{-30±\sqrt{900-50000}}{2\left(-10\right)}$$

Add 900 to -50000.

$$x=\frac{-30±\sqrt{-49100}}{2\left(-10\right)}$$

Take the square root of -49100.

$$x=\frac{-30±10\sqrt{491}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-30±10\sqrt{491}i}{-20}$$

Now solve the equation x=\frac{-30±10\sqrt{491}i}{-20} when ± is plus. Add -30 to 10i\sqrt{491}.

$$x=\frac{-30+10\sqrt{491}i}{-20}$$

Divide -30+10i\sqrt{491} by -20.

$$x=\frac{-\sqrt{491}i+3}{2}$$

Now solve the equation x=\frac{-30±10\sqrt{491}i}{-20} when ± is minus. Subtract 10i\sqrt{491} from -30.

$$x=\frac{-10\sqrt{491}i-30}{-20}$$

Divide -30-10i\sqrt{491} by -20.

$$x=\frac{3+\sqrt{491}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{491}i+3}{2} x=\frac{3+\sqrt{491}i}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$1250=30x-10x^{2}$$

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

$$30x-10x^{2}=1250$$

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.

$$-10x^{2}+30x=1250$$

Divide both sides by -10.

$$\frac{-10x^{2}+30x}{-10}=\frac{1250}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{30}{-10}x=\frac{1250}{-10}$$

Divide 30 by -10.

$$x^{2}-3x=\frac{1250}{-10}$$

Divide 1250 by -10.

$$x^{2}-3x=-125$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-125+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-125+\frac{9}{4}$$

Add -125 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{491}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{491}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{491}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{491}i}{2} x-\frac{3}{2}=-\frac{\sqrt{491}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{491}i}{2} x=\frac{-\sqrt{491}i+3}{2}$$