Solve 20(-x^2-10x+200)=5000

$$20(- { x }^{ 2 } -10x+200)=5000$$

Solve for x (complex solution)

$$x=-5-5i$$

$$x=-5+5i$$

Steps Using the Quadratic Formula

Divide both sides by 20.

$$-x^{2}-10x+200=\frac{5000}{20}$$

Divide 5000 by 20 to get 250.

$$-x^{2}-10x+200=250$$

Subtract 250 from both sides.

$$-x^{2}-10x+200-250=0$$

Subtract 250 from 200 to get -50.

$$-x^{2}-10x-50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and -50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\left(-50\right)}}{2\left(-1\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\left(-50\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\left(-50\right)}}{2\left(-1\right)}$$

Multiply 4 times -50.

$$x=\frac{-\left(-10\right)±\sqrt{100-200}}{2\left(-1\right)}$$

Add 100 to -200.

$$x=\frac{-\left(-10\right)±\sqrt{-100}}{2\left(-1\right)}$$

Take the square root of -100.

$$x=\frac{-\left(-10\right)±10i}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±10i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{10±10i}{-2}$$

Now solve the equation x=\frac{10±10i}{-2} when ± is plus. Add 10 to 10i.

$$x=\frac{10+10i}{-2}$$

Divide 10+10i by -2.

$$x=-5-5i$$

Now solve the equation x=\frac{10±10i}{-2} when ± is minus. Subtract 10i from 10.

$$x=\frac{10-10i}{-2}$$

Divide 10-10i by -2.

$$x=-5+5i$$

The equation is now solved.

$$x=-5-5i x=-5+5i$$

Steps for Completing the Square

Divide both sides by 20.

$$-x^{2}-10x+200=\frac{5000}{20}$$

Divide 5000 by 20 to get 250.

$$-x^{2}-10x+200=250$$

Subtract 200 from both sides.

$$-x^{2}-10x=250-200$$

Subtract 200 from 250 to get 50.

$$-x^{2}-10x=50$$

Divide both sides by -1.

$$\frac{-x^{2}-10x}{-1}=\frac{50}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{10}{-1}\right)x=\frac{50}{-1}$$

Divide -10 by -1.

$$x^{2}+10x=\frac{50}{-1}$$

Divide 50 by -1.

$$x^{2}+10x=-50$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-50+5^{2}$$

Square 5.

$$x^{2}+10x+25=-50+25$$

Add -50 to 25.

$$x^{2}+10x+25=-25$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=-25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{-25}$$

Simplify.

$$x+5=5i x+5=-5i$$

Subtract 5 from both sides of the equation.

$$x=-5+5i x=-5-5i$$