2(1-x^2)+x=0

Solution for 2(1-x^2)+x=0 equation:

2(1-x^2)+x=0

We multiply parentheses

-2x^2+x+2=0

a = -2; b = 1; c = +2;
Δ = b2-4ac
Δ = 12-4·(-2)·2
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{17}}{2*-2}=\frac{-1-\sqrt{17}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*-2}=\frac{-1+\sqrt{17}}{-4}
$