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Solve for x (complex solution)
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x^{2}+10x=-50
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^{2}+10x-\left(-50\right)=-50-\left(-50\right)
Add 50 to both sides of the equation.
x^{2}+10x-\left(-50\right)=0
Subtracting -50 from itself leaves 0.
x^{2}+10x+50=0
Subtract -50 from 0.
x=\frac{-10±\sqrt{10^{2}-4\times 50}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\times 50}}{2}
Square 10.
x=\frac{-10±\sqrt{100-200}}{2}
Multiply -4 times 50.
x=\frac{-10±\sqrt{-100}}{2}
Add 100 to -200.
x=\frac{-10±10i}{2}
Take the square root of -100.
x=\frac{-10+10i}{2}
Now solve the equation x=\frac{-10±10i}{2} when ± is plus. Add -10 to 10i.
x=-5+5i
Divide -10+10i by 2.
x=\frac{-10-10i}{2}
Now solve the equation x=\frac{-10±10i}{2} when ± is minus. Subtract 10i from -10.
x=-5-5i
Divide -10-10i by 2.
x=-5+5i x=-5-5i
The equation is now solved.
x^{2}+10x=-50
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+5^{2}=-50+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=-50+25
Square 5.
x^{2}+10x+25=-25
Add -50 to 25.
\left(x+5\right)^{2}=-25
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{-25}
Take the square root of both sides of the equation.
x+5=5i x+5=-5i
Simplify.
x=-5+5i x=-5-5i
Subtract 5 from both sides of the equation.