4n^2+2n-4=0

Solution for 4n^2+2n-4=0 equation:

4n^2+2n-4=0

a = 4; b = 2; c = -4;
Δ = b2-4ac
Δ = 22-4·4·(-4)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{17}}{2*4}=\frac{-2-2\sqrt{17}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{17}}{2*4}=\frac{-2+2\sqrt{17}}{8}
$