4n^2+2n-2550=0

Solution for 4n^2+2n-2550=0 equation:

4n^2+2n-2550=0

a = 4; b = 2; c = -2550;
Δ = b2-4ac
Δ = 22-4·4·(-2550)
Δ = 40804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{40804}=202$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-202}{2*4}=\frac{-204}{8}
=-25+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+202}{2*4}=\frac{200}{8}
=25
$