Solve 2x^2+40x=-254

$$2 x ^ { 2 } + 40 x = - 254$$

Solve for x (complex solution)

$$x=-10+3\sqrt{3}i\approx -10+5.196152423i$$

$$x=-3\sqrt{3}i-10\approx -10-5.196152423i$$

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.

$$2x^{2}+40x=-254$$

Add 254 to both sides of the equation.

$$2x^{2}+40x-\left(-254\right)=-254-\left(-254\right)$$

Subtracting -254 from itself leaves 0.

$$2x^{2}+40x-\left(-254\right)=0$$

Subtract -254 from 0.

$$2x^{2}+40x+254=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 40 for b, and 254 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 2\times 254}}{2\times 2}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 2\times 254}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-40±\sqrt{1600-8\times 254}}{2\times 2}$$

Multiply -8 times 254.

$$x=\frac{-40±\sqrt{1600-2032}}{2\times 2}$$

Add 1600 to -2032.

$$x=\frac{-40±\sqrt{-432}}{2\times 2}$$

Take the square root of -432.

$$x=\frac{-40±12\sqrt{3}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-40±12\sqrt{3}i}{4}$$

Now solve the equation x=\frac{-40±12\sqrt{3}i}{4} when ± is plus. Add -40 to 12i\sqrt{3}.

$$x=\frac{-40+12\sqrt{3}i}{4}$$

Divide -40+12i\sqrt{3} by 4.

$$x=-10+3\sqrt{3}i$$

Now solve the equation x=\frac{-40±12\sqrt{3}i}{4} when ± is minus. Subtract 12i\sqrt{3} from -40.

$$x=\frac{-12\sqrt{3}i-40}{4}$$

Divide -40-12i\sqrt{3} by 4.

$$x=-3\sqrt{3}i-10$$

The equation is now solved.

$$x=-10+3\sqrt{3}i x=-3\sqrt{3}i-10$$

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.

$$2x^{2}+40x=-254$$

Divide both sides by 2.

$$\frac{2x^{2}+40x}{2}=-\frac{254}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{40}{2}x=-\frac{254}{2}$$

Divide 40 by 2.

$$x^{2}+20x=-\frac{254}{2}$$

Divide -254 by 2.

$$x^{2}+20x=-127$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+20x+10^{2}=-127+10^{2}$$

Square 10.

$$x^{2}+20x+100=-127+100$$

Add -127 to 100.

$$x^{2}+20x+100=-27$$

Factor x^{2}+20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+10\right)^{2}=-27$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{-27}$$

Simplify.

$$x+10=3\sqrt{3}i x+10=-3\sqrt{3}i$$

Subtract 10 from both sides of the equation.

$$x=-10+3\sqrt{3}i x=-3\sqrt{3}i-10$$