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Solve for x (complex solution)
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x^{2}+54x-5=500
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}+54x-5-500=500-500
Subtract 500 from both sides of the equation.
x^{2}+54x-5-500=0
Subtracting 500 from itself leaves 0.
x^{2}+54x-505=0
Subtract 500 from -5.
x=\frac{-54±\sqrt{54^{2}-4\left(-505\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 54 for b, and -505 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-54±\sqrt{2916-4\left(-505\right)}}{2}
Square 54.
x=\frac{-54±\sqrt{2916+2020}}{2}
Multiply -4 times -505.
x=\frac{-54±\sqrt{4936}}{2}
Add 2916 to 2020.
x=\frac{-54±2\sqrt{1234}}{2}
Take the square root of 4936.
x=\frac{2\sqrt{1234}-54}{2}
Now solve the equation x=\frac{-54±2\sqrt{1234}}{2} when ± is plus. Add -54 to 2\sqrt{1234}.
x=\sqrt{1234}-27
Divide -54+2\sqrt{1234} by 2.
x=\frac{-2\sqrt{1234}-54}{2}
Now solve the equation x=\frac{-54±2\sqrt{1234}}{2} when ± is minus. Subtract 2\sqrt{1234} from -54.
x=-\sqrt{1234}-27
Divide -54-2\sqrt{1234} by 2.
x=\sqrt{1234}-27 x=-\sqrt{1234}-27
The equation is now solved.
x^{2}+54x-5=500
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}+54x-5-\left(-5\right)=500-\left(-5\right)
Add 5 to both sides of the equation.
x^{2}+54x=500-\left(-5\right)
Subtracting -5 from itself leaves 0.
x^{2}+54x=505
Subtract -5 from 500.
x^{2}+54x+27^{2}=505+27^{2}
Divide 54, the coefficient of the x term, by 2 to get 27. Then add the square of 27 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+54x+729=505+729
Square 27.
x^{2}+54x+729=1234
Add 505 to 729.
\left(x+27\right)^{2}=1234
Factor x^{2}+54x+729. 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+27\right)^{2}}=\sqrt{1234}
Take the square root of both sides of the equation.
x+27=\sqrt{1234} x+27=-\sqrt{1234}
Simplify.
x=\sqrt{1234}-27 x=-\sqrt{1234}-27
Subtract 27 from both sides of the equation.
x^{2}+54x-5=500
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}+54x-5-500=500-500
Subtract 500 from both sides of the equation.
x^{2}+54x-5-500=0
Subtracting 500 from itself leaves 0.
x^{2}+54x-505=0
Subtract 500 from -5.
x=\frac{-54±\sqrt{54^{2}-4\left(-505\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 54 for b, and -505 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-54±\sqrt{2916-4\left(-505\right)}}{2}
Square 54.
x=\frac{-54±\sqrt{2916+2020}}{2}
Multiply -4 times -505.
x=\frac{-54±\sqrt{4936}}{2}
Add 2916 to 2020.
x=\frac{-54±2\sqrt{1234}}{2}
Take the square root of 4936.
x=\frac{2\sqrt{1234}-54}{2}
Now solve the equation x=\frac{-54±2\sqrt{1234}}{2} when ± is plus. Add -54 to 2\sqrt{1234}.
x=\sqrt{1234}-27
Divide -54+2\sqrt{1234} by 2.
x=\frac{-2\sqrt{1234}-54}{2}
Now solve the equation x=\frac{-54±2\sqrt{1234}}{2} when ± is minus. Subtract 2\sqrt{1234} from -54.
x=-\sqrt{1234}-27
Divide -54-2\sqrt{1234} by 2.
x=\sqrt{1234}-27 x=-\sqrt{1234}-27
The equation is now solved.
x^{2}+54x-5=500
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}+54x-5-\left(-5\right)=500-\left(-5\right)
Add 5 to both sides of the equation.
x^{2}+54x=500-\left(-5\right)
Subtracting -5 from itself leaves 0.
x^{2}+54x=505
Subtract -5 from 500.
x^{2}+54x+27^{2}=505+27^{2}
Divide 54, the coefficient of the x term, by 2 to get 27. Then add the square of 27 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+54x+729=505+729
Square 27.
x^{2}+54x+729=1234
Add 505 to 729.
\left(x+27\right)^{2}=1234
Factor x^{2}+54x+729. 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+27\right)^{2}}=\sqrt{1234}
Take the square root of both sides of the equation.
x+27=\sqrt{1234} x+27=-\sqrt{1234}
Simplify.
x=\sqrt{1234}-27 x=-\sqrt{1234}-27
Subtract 27 from both sides of the equation.