4n^2+16n-6=0

Solution for 4n^2+16n-6=0 equation:

4n^2+16n-6=0

a = 4; b = 16; c = -6;
Δ = b2-4ac
Δ = 162-4·4·(-6)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{22}}{2*4}=\frac{-16-4\sqrt{22}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{22}}{2*4}=\frac{-16+4\sqrt{22}}{8}
$