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Solve for x (complex solution)
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x^{2}+4x=6
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}+4x-6=6-6
Subtract 6 from both sides of the equation.
x^{2}+4x-6=0
Subtracting 6 from itself leaves 0.
x=\frac{-4±\sqrt{4^{2}-4\left(-6\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\left(-6\right)}}{2}
Square 4.
x=\frac{-4±\sqrt{16+24}}{2}
Multiply -4 times -6.
x=\frac{-4±\sqrt{40}}{2}
Add 16 to 24.
x=\frac{-4±2\sqrt{10}}{2}
Take the square root of 40.
x=\frac{2\sqrt{10}-4}{2}
Now solve the equation x=\frac{-4±2\sqrt{10}}{2} when ± is plus. Add -4 to 2\sqrt{10}.
x=\sqrt{10}-2
Divide -4+2\sqrt{10} by 2.
x=\frac{-2\sqrt{10}-4}{2}
Now solve the equation x=\frac{-4±2\sqrt{10}}{2} when ± is minus. Subtract 2\sqrt{10} from -4.
x=-\sqrt{10}-2
Divide -4-2\sqrt{10} by 2.
x=\sqrt{10}-2 x=-\sqrt{10}-2
The equation is now solved.
x^{2}+4x=6
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+2^{2}=6+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=6+4
Square 2.
x^{2}+4x+4=10
Add 6 to 4.
\left(x+2\right)^{2}=10
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{10}
Take the square root of both sides of the equation.
x+2=\sqrt{10} x+2=-\sqrt{10}
Simplify.
x=\sqrt{10}-2 x=-\sqrt{10}-2
Subtract 2 from both sides of the equation.
x^{2}+4x=6
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}+4x-6=6-6
Subtract 6 from both sides of the equation.
x^{2}+4x-6=0
Subtracting 6 from itself leaves 0.
x=\frac{-4±\sqrt{4^{2}-4\left(-6\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\left(-6\right)}}{2}
Square 4.
x=\frac{-4±\sqrt{16+24}}{2}
Multiply -4 times -6.
x=\frac{-4±\sqrt{40}}{2}
Add 16 to 24.
x=\frac{-4±2\sqrt{10}}{2}
Take the square root of 40.
x=\frac{2\sqrt{10}-4}{2}
Now solve the equation x=\frac{-4±2\sqrt{10}}{2} when ± is plus. Add -4 to 2\sqrt{10}.
x=\sqrt{10}-2
Divide -4+2\sqrt{10} by 2.
x=\frac{-2\sqrt{10}-4}{2}
Now solve the equation x=\frac{-4±2\sqrt{10}}{2} when ± is minus. Subtract 2\sqrt{10} from -4.
x=-\sqrt{10}-2
Divide -4-2\sqrt{10} by 2.
x=\sqrt{10}-2 x=-\sqrt{10}-2
The equation is now solved.
x^{2}+4x=6
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+2^{2}=6+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=6+4
Square 2.
x^{2}+4x+4=10
Add 6 to 4.
\left(x+2\right)^{2}=10
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{10}
Take the square root of both sides of the equation.
x+2=\sqrt{10} x+2=-\sqrt{10}
Simplify.
x=\sqrt{10}-2 x=-\sqrt{10}-2
Subtract 2 from both sides of the equation.