(1+4y-2y^2)=0

Solution for (1+4y-2y^2)=0 equation:

(1+4y-2y^2)=0

We get rid of parentheses

-2y^2+4y+1=0

a = -2; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·(-2)·1
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{6}}{2*-2}=\frac{-4-2\sqrt{6}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{6}}{2*-2}=\frac{-4+2\sqrt{6}}{-4}
$