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$2 \exponential{x}{2} + 12 x + 40 = 0 $
Solve for x (complex solution)
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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.