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-2x^{2}+8x+4=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{-8±\sqrt{8^{2}-4\left(-2\right)\times 4}}{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{-8±\sqrt{64-4\left(-2\right)\times 4}}{2\left(-2\right)}
Square 8.
x=\frac{-8±\sqrt{64+8\times 4}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-8±\sqrt{64+32}}{2\left(-2\right)}
Multiply 8 times 4.
x=\frac{-8±\sqrt{96}}{2\left(-2\right)}
x=\frac{-8±4\sqrt{6}}{2\left(-2\right)}
Take the square root of 96.
x=\frac{-8±4\sqrt{6}}{-4}
Multiply 2 times -2.
x=\frac{4\sqrt{6}-8}{-4}
Now solve the equation x=\frac{-8±4\sqrt{6}}{-4} when ± is plus. Add -8 to 4\sqrt{6}.
x=2-\sqrt{6}
Divide -8+4\sqrt{6} by -4.
x=\frac{-4\sqrt{6}-8}{-4}
Now solve the equation x=\frac{-8±4\sqrt{6}}{-4} when ± is minus. Subtract 4\sqrt{6} from -8.
x=\sqrt{6}+2
Divide -8-4\sqrt{6} by -4.
-2x^{2}+8x+4=-2\left(x-\left(2-\sqrt{6}\right)\right)\left(x-\left(\sqrt{6}+2\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2-\sqrt{6} for x_{1} and 2+\sqrt{6} for x_{2}.
x ^ 2 -4x -2 = 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 = 4 rs = -2
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 = 2 - u s = 2 + u
Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 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>
(2 - u) (2 + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
4 - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-4 = -6
Simplify the expression by subtracting 4 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 =2 - \sqrt{6} = -0.449 s = 2 + \sqrt{6} = 4.449
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.