Solve 40x-100-0.5x^2=500

$$40 x - 100 - 0.5 x ^ { 2 } = 500$$

Solve for x

$$x=20$$

$$x=60$$

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.

$$-0.5x^{2}+40x-100=500$$

Subtract 500 from both sides of the equation.

$$-0.5x^{2}+40x-100-500=500-500$$

Subtracting 500 from itself leaves 0.

$$-0.5x^{2}+40x-100-500=0$$

Subtract 500 from -100.

$$-0.5x^{2}+40x-600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.5 for a, 40 for b, and -600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-0.5\right)\left(-600\right)}}{2\left(-0.5\right)}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-0.5\right)\left(-600\right)}}{2\left(-0.5\right)}$$

Multiply -4 times -0.5.

$$x=\frac{-40±\sqrt{1600+2\left(-600\right)}}{2\left(-0.5\right)}$$

Multiply 2 times -600.

$$x=\frac{-40±\sqrt{1600-1200}}{2\left(-0.5\right)}$$

Add 1600 to -1200.

$$x=\frac{-40±\sqrt{400}}{2\left(-0.5\right)}$$

Take the square root of 400.

$$x=\frac{-40±20}{2\left(-0.5\right)}$$

Multiply 2 times -0.5.

$$x=\frac{-40±20}{-1}$$

Now solve the equation x=\frac{-40±20}{-1} when ± is plus. Add -40 to 20.

$$x=-\frac{20}{-1}$$

Divide -20 by -1.

$$x=20$$

Now solve the equation x=\frac{-40±20}{-1} when ± is minus. Subtract 20 from -40.

$$x=-\frac{60}{-1}$$

Divide -60 by -1.

$$x=60$$

The equation is now solved.

$$x=20 x=60$$

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.

$$-0.5x^{2}+40x-100=500$$

Add 100 to both sides of the equation.

$$-0.5x^{2}+40x-100-\left(-100\right)=500-\left(-100\right)$$

Subtracting -100 from itself leaves 0.

$$-0.5x^{2}+40x=500-\left(-100\right)$$

Subtract -100 from 500.

$$-0.5x^{2}+40x=600$$

Multiply both sides by -2.

$$\frac{-0.5x^{2}+40x}{-0.5}=\frac{600}{-0.5}$$

Dividing by -0.5 undoes the multiplication by -0.5.

$$x^{2}+\frac{40}{-0.5}x=\frac{600}{-0.5}$$

Divide 40 by -0.5 by multiplying 40 by the reciprocal of -0.5.

$$x^{2}-80x=\frac{600}{-0.5}$$

Divide 600 by -0.5 by multiplying 600 by the reciprocal of -0.5.

$$x^{2}-80x=-1200$$

Divide -80, the coefficient of the x term, by 2 to get -40. Then add the square of -40 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-80x+\left(-40\right)^{2}=-1200+\left(-40\right)^{2}$$

Square -40.

$$x^{2}-80x+1600=-1200+1600$$

Add -1200 to 1600.

$$x^{2}-80x+1600=400$$

Factor x^{2}-80x+1600. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-40\right)^{2}=400$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-40\right)^{2}}=\sqrt{400}$$

Simplify.

$$x-40=20 x-40=-20$$

Add 40 to both sides of the equation.

$$x=60 x=20$$