49n^2+41n=0

Solution for 49n^2+41n=0 equation:

49n^2+41n=0

a = 49; b = 41; c = 0;
Δ = b2-4ac
Δ = 412-4·49·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-41}{2*49}=\frac{-82}{98}
=-41/49
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+41}{2*49}=\frac{0}{98}
=0
$