Solve 40x+x^2=800

$$40 x + x ^ { 2 } = 800$$

Solve for x

$$x=20\sqrt{3}-20\approx 14.641016151$$

$$x=-20\sqrt{3}-20\approx -54.641016151$$

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.

$$x^{2}+40x=800$$

Subtract 800 from both sides of the equation.

$$x^{2}+40x-800=800-800$$

Subtracting 800 from itself leaves 0.

$$x^{2}+40x-800=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 40 for b, and -800 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-800\right)}}{2}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-800\right)}}{2}$$

Multiply -4 times -800.

$$x=\frac{-40±\sqrt{1600+3200}}{2}$$

Add 1600 to 3200.

$$x=\frac{-40±\sqrt{4800}}{2}$$

Take the square root of 4800.

$$x=\frac{-40±40\sqrt{3}}{2}$$

Now solve the equation x=\frac{-40±40\sqrt{3}}{2} when ± is plus. Add -40 to 40\sqrt{3}.

$$x=\frac{40\sqrt{3}-40}{2}$$

Divide -40+40\sqrt{3} by 2.

$$x=20\sqrt{3}-20$$

Now solve the equation x=\frac{-40±40\sqrt{3}}{2} when ± is minus. Subtract 40\sqrt{3} from -40.

$$x=\frac{-40\sqrt{3}-40}{2}$$

Divide -40-40\sqrt{3} by 2.

$$x=-20\sqrt{3}-20$$

The equation is now solved.

$$x=20\sqrt{3}-20 x=-20\sqrt{3}-20$$

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.

$$x^{2}+40x=800$$

Divide 40, the coefficient of the x term, by 2 to get 20. Then add the square of 20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+40x+20^{2}=800+20^{2}$$

Square 20.

$$x^{2}+40x+400=800+400$$

Add 800 to 400.

$$x^{2}+40x+400=1200$$

Factor x^{2}+40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+20\right)^{2}=1200$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+20\right)^{2}}=\sqrt{1200}$$

Simplify.

$$x+20=20\sqrt{3} x+20=-20\sqrt{3}$$

Subtract 20 from both sides of the equation.

$$x=20\sqrt{3}-20 x=-20\sqrt{3}-20$$