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Solve for x (complex solution)
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x^{2}+14x+50=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{-14±\sqrt{14^{2}-4\times 50}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 14 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-14±\sqrt{196-4\times 50}}{2}
Square 14.
x=\frac{-14±\sqrt{196-200}}{2}
Multiply -4 times 50.
x=\frac{-14±\sqrt{-4}}{2}
Add 196 to -200.
x=\frac{-14±2i}{2}
Take the square root of -4.
x=\frac{-14+2i}{2}
Now solve the equation x=\frac{-14±2i}{2} when ± is plus. Add -14 to 2i.
x=-7+i
Divide -14+2i by 2.
x=\frac{-14-2i}{2}
Now solve the equation x=\frac{-14±2i}{2} when ± is minus. Subtract 2i from -14.
x=-7-i
Divide -14-2i by 2.
x=-7+i x=-7-i
The equation is now solved.
x^{2}+14x+50=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}+14x+50-50=-50
Subtract 50 from both sides of the equation.
x^{2}+14x=-50
Subtracting 50 from itself leaves 0.
x^{2}+14x+7^{2}=-50+7^{2}
Divide 14, the coefficient of the x term, by 2 to get 7. Then add the square of 7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+14x+49=-50+49
Square 7.
x^{2}+14x+49=-1
Add -50 to 49.
\left(x+7\right)^{2}=-1
Factor x^{2}+14x+49. 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+7\right)^{2}}=\sqrt{-1}
Take the square root of both sides of the equation.
x+7=i x+7=-i
Simplify.
x=-7+i x=-7-i
Subtract 7 from both sides of the equation.