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2\left(-x^{2}-x-1\right)
Factor out 2. Polynomial -x^{2}-x-1 is not factored since it does not have any rational roots.
-2x^{2}-2x-2=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}
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{4-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4+8\left(-2\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-\left(-2\right)±\sqrt{4-16}}{2\left(-2\right)}
Multiply 8 times -2.
x=\frac{-\left(-2\right)±\sqrt{-12}}{2\left(-2\right)}
Add 4 to -16.
-2x^{2}-2x-2
Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.
x ^ 2 +1x +1 = 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 = -1 rs = 1
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 = -\frac{1}{2} - u s = -\frac{1}{2} + u
Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. 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>
(-\frac{1}{2} - u) (-\frac{1}{2} + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{1}{4} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{1}{4} = \frac{3}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -\frac{3}{4} u = \pm\sqrt{-\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{1}{2} - \frac{\sqrt{3}}{2}i = -0.500 - 0.866i s = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = -0.500 + 0.866i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.