35n^2+2n-48=0

Solution for 35n^2+2n-48=0 equation:

35n^2+2n-48=0

a = 35; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·35·(-48)
Δ = 6724
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6724}=82$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-82}{2*35}=\frac{-84}{70}
=-1+1/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+82}{2*35}=\frac{80}{70}
=1+1/7
$