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Solve for x (complex solution)
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x^{2}+5x+7=0
Combine 3x and 2x to get 5x.
x=\frac{-5±\sqrt{5^{2}-4\times 7}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\times 7}}{2}
Square 5.
x=\frac{-5±\sqrt{25-28}}{2}
Multiply -4 times 7.
x=\frac{-5±\sqrt{-3}}{2}
Add 25 to -28.
x=\frac{-5±\sqrt{3}i}{2}
Take the square root of -3.
x=\frac{-5+\sqrt{3}i}{2}
Now solve the equation x=\frac{-5±\sqrt{3}i}{2} when ± is plus. Add -5 to i\sqrt{3}.
x=\frac{-\sqrt{3}i-5}{2}
Now solve the equation x=\frac{-5±\sqrt{3}i}{2} when ± is minus. Subtract i\sqrt{3} from -5.
x=\frac{-5+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-5}{2}
The equation is now solved.
x^{2}+5x+7=0
Combine 3x and 2x to get 5x.
x^{2}+5x=-7
Subtract 7 from both sides. Anything subtracted from zero gives its negation.
x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-7+\left(\frac{5}{2}\right)^{2}
Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+5x+\frac{25}{4}=-7+\frac{25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+5x+\frac{25}{4}=-\frac{3}{4}
Add -7 to \frac{25}{4}.
\left(x+\frac{5}{2}\right)^{2}=-\frac{3}{4}
Factor x^{2}+5x+\frac{25}{4}. 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+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{3}{4}}
Take the square root of both sides of the equation.
x+\frac{5}{2}=\frac{\sqrt{3}i}{2} x+\frac{5}{2}=-\frac{\sqrt{3}i}{2}
Simplify.
x=\frac{-5+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-5}{2}
Subtract \frac{5}{2} from both sides of the equation.