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Solve for x (complex solution)
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x^{2}+8x+5=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{-8±\sqrt{8^{2}-4\times 5}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-8±\sqrt{64-4\times 5}}{2}
Square 8.
x=\frac{-8±\sqrt{64-20}}{2}
Multiply -4 times 5.
x=\frac{-8±\sqrt{44}}{2}
Add 64 to -20.
x=\frac{-8±2\sqrt{11}}{2}
Take the square root of 44.
x=\frac{2\sqrt{11}-8}{2}
Now solve the equation x=\frac{-8±2\sqrt{11}}{2} when ± is plus. Add -8 to 2\sqrt{11}.
x=\sqrt{11}-4
Divide -8+2\sqrt{11} by 2.
x=\frac{-2\sqrt{11}-8}{2}
Now solve the equation x=\frac{-8±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from -8.
x=-\sqrt{11}-4
Divide -8-2\sqrt{11} by 2.
x=\sqrt{11}-4 x=-\sqrt{11}-4
The equation is now solved.
x^{2}+8x+5=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}+8x+5-5=-5
Subtract 5 from both sides of the equation.
x^{2}+8x=-5
Subtracting 5 from itself leaves 0.
x^{2}+8x+4^{2}=-5+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+8x+16=-5+16
Square 4.
x^{2}+8x+16=11
Add -5 to 16.
\left(x+4\right)^{2}=11
Factor x^{2}+8x+16. 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+4\right)^{2}}=\sqrt{11}
Take the square root of both sides of the equation.
x+4=\sqrt{11} x+4=-\sqrt{11}
Simplify.
x=\sqrt{11}-4 x=-\sqrt{11}-4
Subtract 4 from both sides of the equation.
x^{2}+8x+5=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{-8±\sqrt{8^{2}-4\times 5}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-8±\sqrt{64-4\times 5}}{2}
Square 8.
x=\frac{-8±\sqrt{64-20}}{2}
Multiply -4 times 5.
x=\frac{-8±\sqrt{44}}{2}
Add 64 to -20.
x=\frac{-8±2\sqrt{11}}{2}
Take the square root of 44.
x=\frac{2\sqrt{11}-8}{2}
Now solve the equation x=\frac{-8±2\sqrt{11}}{2} when ± is plus. Add -8 to 2\sqrt{11}.
x=\sqrt{11}-4
Divide -8+2\sqrt{11} by 2.
x=\frac{-2\sqrt{11}-8}{2}
Now solve the equation x=\frac{-8±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from -8.
x=-\sqrt{11}-4
Divide -8-2\sqrt{11} by 2.
x=\sqrt{11}-4 x=-\sqrt{11}-4
The equation is now solved.
x^{2}+8x+5=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}+8x+5-5=-5
Subtract 5 from both sides of the equation.
x^{2}+8x=-5
Subtracting 5 from itself leaves 0.
x^{2}+8x+4^{2}=-5+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+8x+16=-5+16
Square 4.
x^{2}+8x+16=11
Add -5 to 16.
\left(x+4\right)^{2}=11
Factor x^{2}+8x+16. 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+4\right)^{2}}=\sqrt{11}
Take the square root of both sides of the equation.
x+4=\sqrt{11} x+4=-\sqrt{11}
Simplify.
x=\sqrt{11}-4 x=-\sqrt{11}-4
Subtract 4 from both sides of the equation.