1.75n^2+28.25n+105=0

Solution for 1.75n^2+28.25n+105=0 equation:

1.75n^2+28.25n+105=0

a = 1.75; b = 28.25; c = +105;
Δ = b2-4ac
Δ = 28.252-4·1.75·105
Δ = 63.0625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28.25)-\sqrt{63.0625}}{2*1.75}=\frac{-28.25-\sqrt{63.0625}}{3.5}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28.25)+\sqrt{63.0625}}{2*1.75}=\frac{-28.25+\sqrt{63.0625}}{3.5}
$