40n^2+48n+8=0

Solution for 40n^2+48n+8=0 equation:

40n^2+48n+8=0

a = 40; b = 48; c = +8;
Δ = b2-4ac
Δ = 482-4·40·8
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-32}{2*40}=\frac{-80}{80}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+32}{2*40}=\frac{-16}{80}
=-1/5
$