9n^2+42n+35=0

Solution for 9n^2+42n+35=0 equation:

9n^2+42n+35=0

a = 9; b = 42; c = +35;
Δ = b2-4ac
Δ = 422-4·9·35
Δ = 504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{14}}{2*9}=\frac{-42-6\sqrt{14}}{18}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{14}}{2*9}=\frac{-42+6\sqrt{14}}{18}
$