Solve for x (complex solution)

x=-3+\sqrt{11}i\approx -3+3.31662479i<br/>x=-\sqrt{11}i-3\approx -3-3.31662479i

$x=−3+11 i≈−3+3.31662479i$

$x=−11 i−3≈−3−3.31662479i$

$x=−11 i−3≈−3−3.31662479i$

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2x^{2}+12x+40=0

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=\frac{-12±\sqrt{12^{2}-4\times 2\times 40}}{2\times 2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 12 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-12±\sqrt{144-4\times 2\times 40}}{2\times 2}

Square 12.

x=\frac{-12±\sqrt{144-8\times 40}}{2\times 2}

Multiply -4 times 2.

x=\frac{-12±\sqrt{144-320}}{2\times 2}

Multiply -8 times 40.

x=\frac{-12±\sqrt{-176}}{2\times 2}

Add 144 to -320.

x=\frac{-12±4\sqrt{11}i}{2\times 2}

Take the square root of -176.

x=\frac{-12±4\sqrt{11}i}{4}

Multiply 2 times 2.

x=\frac{-12+4\sqrt{11}i}{4}

Now solve the equation x=\frac{-12±4\sqrt{11}i}{4} when ± is plus. Add -12 to 4i\sqrt{11}.

x=-3+\sqrt{11}i

Divide -12+4i\sqrt{11} by 4.

x=\frac{-4\sqrt{11}i-12}{4}

Now solve the equation x=\frac{-12±4\sqrt{11}i}{4} when ± is minus. Subtract 4i\sqrt{11} from -12.

x=-\sqrt{11}i-3

Divide -12-4i\sqrt{11} by 4.

x=-3+\sqrt{11}i x=-\sqrt{11}i-3

The equation is now solved.

2x^{2}+12x+40=0

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}+12x+40-40=-40

Subtract 40 from both sides of the equation.

2x^{2}+12x=-40

Subtracting 40 from itself leaves 0.

\frac{2x^{2}+12x}{2}=\frac{-40}{2}

Divide both sides by 2.

x^{2}+\frac{12}{2}x=\frac{-40}{2}

Dividing by 2 undoes the multiplication by 2.

x^{2}+6x=\frac{-40}{2}

Divide 12 by 2.

x^{2}+6x=-20

Divide -40 by 2.

x^{2}+6x+3^{2}=-20+3^{2}

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}+6x+9=-20+9

Square 3.

x^{2}+6x+9=-11

Add -20 to 9.

\left(x+3\right)^{2}=-11

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

\sqrt{\left(x+3\right)^{2}}=\sqrt{-11}

Take the square root of both sides of the equation.

x+3=\sqrt{11}i x+3=-\sqrt{11}i

Simplify.

x=-3+\sqrt{11}i x=-\sqrt{11}i-3

Subtract 3 from both sides of the equation.

x ^ 2 +6x +20 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 2

r + s = -6 rs = 20

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = -3 - u s = -3 + u

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(-3 - u) (-3 + u) = 20

To solve for unknown quantity u, substitute these in the product equation rs = 20

9 - u^2 = 20

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = 20-9 = 11

Simplify the expression by subtracting 9 on both sides

u^2 = -11 u = \pm\sqrt{-11} = \pm \sqrt{11}i

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =-3 - \sqrt{11}i s = -3 + \sqrt{11}i

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.