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Solve for x (complex solution)
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x^{2}+16x+128=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{-16±\sqrt{16^{2}-4\times 128}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 16 for b, and 128 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-16±\sqrt{256-4\times 128}}{2}
Square 16.
x=\frac{-16±\sqrt{256-512}}{2}
Multiply -4 times 128.
x=\frac{-16±\sqrt{-256}}{2}
Add 256 to -512.
x=\frac{-16±16i}{2}
Take the square root of -256.
x=\frac{-16+16i}{2}
Now solve the equation x=\frac{-16±16i}{2} when ± is plus. Add -16 to 16i.
x=-8+8i
Divide -16+16i by 2.
x=\frac{-16-16i}{2}
Now solve the equation x=\frac{-16±16i}{2} when ± is minus. Subtract 16i from -16.
x=-8-8i
Divide -16-16i by 2.
x=-8+8i x=-8-8i
The equation is now solved.
x^{2}+16x+128=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}+16x+128-128=-128
Subtract 128 from both sides of the equation.
x^{2}+16x=-128
Subtracting 128 from itself leaves 0.
x^{2}+16x+8^{2}=-128+8^{2}
Divide 16, the coefficient of the x term, by 2 to get 8. Then add the square of 8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+16x+64=-128+64
Square 8.
x^{2}+16x+64=-64
Add -128 to 64.
\left(x+8\right)^{2}=-64
Factor x^{2}+16x+64. 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+8\right)^{2}}=\sqrt{-64}
Take the square root of both sides of the equation.
x+8=8i x+8=-8i
Simplify.
x=-8+8i x=-8-8i
Subtract 8 from both sides of the equation.