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Solve for x (complex solution)
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x^{2}+10x+29=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{-10±\sqrt{10^{2}-4\times 29}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\times 29}}{2}
Square 10.
x=\frac{-10±\sqrt{100-116}}{2}
Multiply -4 times 29.
x=\frac{-10±\sqrt{-16}}{2}
Add 100 to -116.
x=\frac{-10±4i}{2}
Take the square root of -16.
x=\frac{-10+4i}{2}
Now solve the equation x=\frac{-10±4i}{2} when ± is plus. Add -10 to 4i.
x=-5+2i
Divide -10+4i by 2.
x=\frac{-10-4i}{2}
Now solve the equation x=\frac{-10±4i}{2} when ± is minus. Subtract 4i from -10.
x=-5-2i
Divide -10-4i by 2.
x=-5+2i x=-5-2i
The equation is now solved.
x^{2}+10x+29=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.
x^{2}+10x+29-29=-29
Subtract 29 from both sides of the equation.
x^{2}+10x=-29
Subtracting 29 from itself leaves 0.
x^{2}+10x+5^{2}=-29+5^{2}
Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+10x+25=-29+25
Square 5.
x^{2}+10x+25=-4
Add -29 to 25.
\left(x+5\right)^{2}=-4
Factor x^{2}+10x+25. 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+5\right)^{2}}=\sqrt{-4}
Take the square root of both sides of the equation.
x+5=2i x+5=-2i
Simplify.
x=-5+2i x=-5-2i
Subtract 5 from both sides of the equation.