2n^2-29n-2040=0

Solution for 2n^2-29n-2040=0 equation:

2n^2-29n-2040=0

a = 2; b = -29; c = -2040;
Δ = b2-4ac
Δ = -292-4·2·(-2040)
Δ = 17161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{17161}=131$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-131}{2*2}=\frac{-102}{4}
=-25+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+131}{2*2}=\frac{160}{4}
=40
$