4n^2+28n-12=0

Solution for 4n^2+28n-12=0 equation:

4n^2+28n-12=0

a = 4; b = 28; c = -12;
Δ = b2-4ac
Δ = 282-4·4·(-12)
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{61}}{2*4}=\frac{-28-4\sqrt{61}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{61}}{2*4}=\frac{-28+4\sqrt{61}}{8}
$