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Solve for x (complex solution)
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-x^{2}-2x+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{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 5}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 5}}{2\left(-1\right)}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4+4\times 5}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-2\right)±\sqrt{4+20}}{2\left(-1\right)}
Multiply 4 times 5.
x=\frac{-\left(-2\right)±\sqrt{24}}{2\left(-1\right)}
Add 4 to 20.
x=\frac{-\left(-2\right)±2\sqrt{6}}{2\left(-1\right)}
Take the square root of 24.
x=\frac{2±2\sqrt{6}}{2\left(-1\right)}
The opposite of -2 is 2.
x=\frac{2±2\sqrt{6}}{-2}
Multiply 2 times -1.
x=\frac{2\sqrt{6}+2}{-2}
Now solve the equation x=\frac{2±2\sqrt{6}}{-2} when ± is plus. Add 2 to 2\sqrt{6}.
x=-\left(\sqrt{6}+1\right)
Divide 2+2\sqrt{6} by -2.
x=\frac{2-2\sqrt{6}}{-2}
Now solve the equation x=\frac{2±2\sqrt{6}}{-2} when ± is minus. Subtract 2\sqrt{6} from 2.
x=\sqrt{6}-1
Divide 2-2\sqrt{6} by -2.
x=-\left(\sqrt{6}+1\right) x=\sqrt{6}-1
The equation is now solved.
-x^{2}-2x+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}-2x+5-5=-5
Subtract 5 from both sides of the equation.
-x^{2}-2x=-5
Subtracting 5 from itself leaves 0.
\frac{-x^{2}-2x}{-1}=-\frac{5}{-1}
Divide both sides by -1.
x^{2}+\left(-\frac{2}{-1}\right)x=-\frac{5}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}+2x=-\frac{5}{-1}
Divide -2 by -1.
x^{2}+2x=5
Divide -5 by -1.
x^{2}+2x+1^{2}=5+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=5+1
Square 1.
x^{2}+2x+1=6
Add 5 to 1.
\left(x+1\right)^{2}=6
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{6}
Take the square root of both sides of the equation.
x+1=\sqrt{6} x+1=-\sqrt{6}
Simplify.
x=\sqrt{6}-1 x=-\sqrt{6}-1
Subtract 1 from both sides of the equation.
x ^ 2 +2x -5 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = -2 rs = -5
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -1 - u s = -1 + u
Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-1 - u) (-1 + u) = -5
To solve for unknown quantity u, substitute these in the product equation rs = -5
1 - u^2 = -5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -5-1 = -6
Simplify the expression by subtracting 1 on both sides
u^2 = 6 u = \pm\sqrt{6} = \pm \sqrt{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-1 - \sqrt{6} = -3.449 s = -1 + \sqrt{6} = 1.449
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
-x^{2}-2x+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{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 5}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 5}}{2\left(-1\right)}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4+4\times 5}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-2\right)±\sqrt{4+20}}{2\left(-1\right)}
Multiply 4 times 5.
x=\frac{-\left(-2\right)±\sqrt{24}}{2\left(-1\right)}
Add 4 to 20.
x=\frac{-\left(-2\right)±2\sqrt{6}}{2\left(-1\right)}
Take the square root of 24.
x=\frac{2±2\sqrt{6}}{2\left(-1\right)}
The opposite of -2 is 2.
x=\frac{2±2\sqrt{6}}{-2}
Multiply 2 times -1.
x=\frac{2\sqrt{6}+2}{-2}
Now solve the equation x=\frac{2±2\sqrt{6}}{-2} when ± is plus. Add 2 to 2\sqrt{6}.
x=-\left(\sqrt{6}+1\right)
Divide 2+2\sqrt{6} by -2.
x=\frac{2-2\sqrt{6}}{-2}
Now solve the equation x=\frac{2±2\sqrt{6}}{-2} when ± is minus. Subtract 2\sqrt{6} from 2.
x=\sqrt{6}-1
Divide 2-2\sqrt{6} by -2.
x=-\left(\sqrt{6}+1\right) x=\sqrt{6}-1
The equation is now solved.
-x^{2}-2x+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}-2x+5-5=-5
Subtract 5 from both sides of the equation.
-x^{2}-2x=-5
Subtracting 5 from itself leaves 0.
\frac{-x^{2}-2x}{-1}=-\frac{5}{-1}
Divide both sides by -1.
x^{2}+\left(-\frac{2}{-1}\right)x=-\frac{5}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}+2x=-\frac{5}{-1}
Divide -2 by -1.
x^{2}+2x=5
Divide -5 by -1.
x^{2}+2x+1^{2}=5+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=5+1
Square 1.
x^{2}+2x+1=6
Add 5 to 1.
\left(x+1\right)^{2}=6
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{6}
Take the square root of both sides of the equation.
x+1=\sqrt{6} x+1=-\sqrt{6}
Simplify.
x=\sqrt{6}-1 x=-\sqrt{6}-1
Subtract 1 from both sides of the equation.