Solve x^2+40x-150=0

$${ x }^{ 2 } +40x-150=0$$

Solve for x

$$x=5\sqrt{22}-20\approx 3.452078799$$

$$x=-5\sqrt{22}-20\approx -43.452078799$$

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.

$$x^{2}+40x-150=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 40 for b, and -150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-150\right)}}{2}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-150\right)}}{2}$$

Multiply -4 times -150.

$$x=\frac{-40±\sqrt{1600+600}}{2}$$

Add 1600 to 600.

$$x=\frac{-40±\sqrt{2200}}{2}$$

Take the square root of 2200.

$$x=\frac{-40±10\sqrt{22}}{2}$$

Now solve the equation x=\frac{-40±10\sqrt{22}}{2} when ± is plus. Add -40 to 10\sqrt{22}.

$$x=\frac{10\sqrt{22}-40}{2}$$

Divide -40+10\sqrt{22} by 2.

$$x=5\sqrt{22}-20$$

Now solve the equation x=\frac{-40±10\sqrt{22}}{2} when ± is minus. Subtract 10\sqrt{22} from -40.

$$x=\frac{-10\sqrt{22}-40}{2}$$

Divide -40-10\sqrt{22} by 2.

$$x=-5\sqrt{22}-20$$

The equation is now solved.

$$x=5\sqrt{22}-20 x=-5\sqrt{22}-20$$

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.

$$x^{2}+40x-150=0$$

Add 150 to both sides of the equation.

$$x^{2}+40x-150-\left(-150\right)=-\left(-150\right)$$

Subtracting -150 from itself leaves 0.

$$x^{2}+40x=-\left(-150\right)$$

Subtract -150 from 0.

$$x^{2}+40x=150$$

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.

$$x^{2}+40x+20^{2}=150+20^{2}$$

Square 20.

$$x^{2}+40x+400=150+400$$

Add 150 to 400.

$$x^{2}+40x+400=550$$

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}.

$$\left(x+20\right)^{2}=550$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+20\right)^{2}}=\sqrt{550}$$

Simplify.

$$x+20=5\sqrt{22} x+20=-5\sqrt{22}$$

Subtract 20 from both sides of the equation.

$$x=5\sqrt{22}-20 x=-5\sqrt{22}-20$$