4n^2+24n-448=0

Solution for 4n^2+24n-448=0 equation:

4n^2+24n-448=0

a = 4; b = 24; c = -448;
Δ = b2-4ac
Δ = 242-4·4·(-448)
Δ = 7744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7744}=88$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-88}{2*4}=\frac{-112}{8}
=-14
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+88}{2*4}=\frac{64}{8}
=8
$