Solve for x (complex solution)
x=\sqrt{3}-1\approx 0.732050808
x=-\left(\sqrt{3}+1\right)\approx -2.732050808
Solve for x
x=\sqrt{3}-1\approx 0.732050808
x=-\sqrt{3}-1\approx -2.732050808
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x^{2}+4-4x+x^{2}=\left(2x\right)^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.
2x^{2}+4-4x=\left(2x\right)^{2}
Combine x^{2} and x^{2} to get 2x^{2}.
2x^{2}+4-4x=2^{2}x^{2}
Expand \left(2x\right)^{2}.
2x^{2}+4-4x=4x^{2}
Calculate 2 to the power of 2 and get 4.
2x^{2}+4-4x-4x^{2}=0
Subtract 4x^{2} from both sides.
-2x^{2}+4-4x=0
Combine 2x^{2} and -4x^{2} to get -2x^{2}.
-2x^{2}-4x+4=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{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-2\right)\times 4}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -4 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 4}}{2\left(-2\right)}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16+8\times 4}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\left(-2\right)}
Multiply 8 times 4.
x=\frac{-\left(-4\right)±\sqrt{48}}{2\left(-2\right)}
Add 16 to 32.
x=\frac{-\left(-4\right)±4\sqrt{3}}{2\left(-2\right)}
Take the square root of 48.
x=\frac{4±4\sqrt{3}}{2\left(-2\right)}
The opposite of -4 is 4.
x=\frac{4±4\sqrt{3}}{-4}
Multiply 2 times -2.
x=\frac{4\sqrt{3}+4}{-4}
Now solve the equation x=\frac{4±4\sqrt{3}}{-4} when ± is plus. Add 4 to 4\sqrt{3}.
x=-\left(\sqrt{3}+1\right)
Divide 4+4\sqrt{3} by -4.
x=\frac{4-4\sqrt{3}}{-4}
Now solve the equation x=\frac{4±4\sqrt{3}}{-4} when ± is minus. Subtract 4\sqrt{3} from 4.
x=\sqrt{3}-1
Divide 4-4\sqrt{3} by -4.
x=-\left(\sqrt{3}+1\right) x=\sqrt{3}-1
The equation is now solved.
x^{2}+4-4x+x^{2}=\left(2x\right)^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.
2x^{2}+4-4x=\left(2x\right)^{2}
Combine x^{2} and x^{2} to get 2x^{2}.
2x^{2}+4-4x=2^{2}x^{2}
Expand \left(2x\right)^{2}.
2x^{2}+4-4x=4x^{2}
Calculate 2 to the power of 2 and get 4.
2x^{2}+4-4x-4x^{2}=0
Subtract 4x^{2} from both sides.
-2x^{2}+4-4x=0
Combine 2x^{2} and -4x^{2} to get -2x^{2}.
-2x^{2}-4x=-4
Subtract 4 from both sides. Anything subtracted from zero gives its negation.
\frac{-2x^{2}-4x}{-2}=-\frac{4}{-2}
Divide both sides by -2.
x^{2}+\left(-\frac{4}{-2}\right)x=-\frac{4}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}+2x=-\frac{4}{-2}
Divide -4 by -2.
x^{2}+2x=2
Divide -4 by -2.
x^{2}+2x+1^{2}=2+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=2+1
Square 1.
x^{2}+2x+1=3
Add 2 to 1.
\left(x+1\right)^{2}=3
Factor x^{2}+2x+1. 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+1\right)^{2}}=\sqrt{3}
Take the square root of both sides of the equation.
x+1=\sqrt{3} x+1=-\sqrt{3}
Simplify.
x=\sqrt{3}-1 x=-\sqrt{3}-1
Subtract 1 from both sides of the equation.
x^{2}+4-4x+x^{2}=\left(2x\right)^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.
2x^{2}+4-4x=\left(2x\right)^{2}
Combine x^{2} and x^{2} to get 2x^{2}.
2x^{2}+4-4x=2^{2}x^{2}
Expand \left(2x\right)^{2}.
2x^{2}+4-4x=4x^{2}
Calculate 2 to the power of 2 and get 4.
2x^{2}+4-4x-4x^{2}=0
Subtract 4x^{2} from both sides.
-2x^{2}+4-4x=0
Combine 2x^{2} and -4x^{2} to get -2x^{2}.
-2x^{2}-4x+4=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{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-2\right)\times 4}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -4 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 4}}{2\left(-2\right)}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16+8\times 4}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\left(-2\right)}
Multiply 8 times 4.
x=\frac{-\left(-4\right)±\sqrt{48}}{2\left(-2\right)}
Add 16 to 32.
x=\frac{-\left(-4\right)±4\sqrt{3}}{2\left(-2\right)}
Take the square root of 48.
x=\frac{4±4\sqrt{3}}{2\left(-2\right)}
The opposite of -4 is 4.
x=\frac{4±4\sqrt{3}}{-4}
Multiply 2 times -2.
x=\frac{4\sqrt{3}+4}{-4}
Now solve the equation x=\frac{4±4\sqrt{3}}{-4} when ± is plus. Add 4 to 4\sqrt{3}.
x=-\left(\sqrt{3}+1\right)
Divide 4+4\sqrt{3} by -4.
x=\frac{4-4\sqrt{3}}{-4}
Now solve the equation x=\frac{4±4\sqrt{3}}{-4} when ± is minus. Subtract 4\sqrt{3} from 4.
x=\sqrt{3}-1
Divide 4-4\sqrt{3} by -4.
x=-\left(\sqrt{3}+1\right) x=\sqrt{3}-1
The equation is now solved.
x^{2}+4-4x+x^{2}=\left(2x\right)^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.
2x^{2}+4-4x=\left(2x\right)^{2}
Combine x^{2} and x^{2} to get 2x^{2}.
2x^{2}+4-4x=2^{2}x^{2}
Expand \left(2x\right)^{2}.
2x^{2}+4-4x=4x^{2}
Calculate 2 to the power of 2 and get 4.
2x^{2}+4-4x-4x^{2}=0
Subtract 4x^{2} from both sides.
-2x^{2}+4-4x=0
Combine 2x^{2} and -4x^{2} to get -2x^{2}.
-2x^{2}-4x=-4
Subtract 4 from both sides. Anything subtracted from zero gives its negation.
\frac{-2x^{2}-4x}{-2}=-\frac{4}{-2}
Divide both sides by -2.
x^{2}+\left(-\frac{4}{-2}\right)x=-\frac{4}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}+2x=-\frac{4}{-2}
Divide -4 by -2.
x^{2}+2x=2
Divide -4 by -2.
x^{2}+2x+1^{2}=2+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=2+1
Square 1.
x^{2}+2x+1=3
Add 2 to 1.
\left(x+1\right)^{2}=3
Factor x^{2}+2x+1. 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+1\right)^{2}}=\sqrt{3}
Take the square root of both sides of the equation.
x+1=\sqrt{3} x+1=-\sqrt{3}
Simplify.
x=\sqrt{3}-1 x=-\sqrt{3}-1
Subtract 1 from both sides of the equation.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}