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Solve for x (complex solution)
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x^{2}+4x+145=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{-4±\sqrt{4^{2}-4\times 145}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and 145 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\times 145}}{2}
Square 4.
x=\frac{-4±\sqrt{16-580}}{2}
Multiply -4 times 145.
x=\frac{-4±\sqrt{-564}}{2}
Add 16 to -580.
x=\frac{-4±2\sqrt{141}i}{2}
Take the square root of -564.
x=\frac{-4+2\sqrt{141}i}{2}
Now solve the equation x=\frac{-4±2\sqrt{141}i}{2} when ± is plus. Add -4 to 2i\sqrt{141}.
x=-2+\sqrt{141}i
Divide -4+2i\sqrt{141} by 2.
x=\frac{-2\sqrt{141}i-4}{2}
Now solve the equation x=\frac{-4±2\sqrt{141}i}{2} when ± is minus. Subtract 2i\sqrt{141} from -4.
x=-\sqrt{141}i-2
Divide -4-2i\sqrt{141} by 2.
x=-2+\sqrt{141}i x=-\sqrt{141}i-2
The equation is now solved.
x^{2}+4x+145=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}+4x+145-145=-145
Subtract 145 from both sides of the equation.
x^{2}+4x=-145
Subtracting 145 from itself leaves 0.
x^{2}+4x+2^{2}=-145+2^{2}
Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+4x+4=-145+4
Square 2.
x^{2}+4x+4=-141
Add -145 to 4.
\left(x+2\right)^{2}=-141
Factor x^{2}+4x+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+2\right)^{2}}=\sqrt{-141}
Take the square root of both sides of the equation.
x+2=\sqrt{141}i x+2=-\sqrt{141}i
Simplify.
x=-2+\sqrt{141}i x=-\sqrt{141}i-2
Subtract 2 from both sides of the equation.