7n^2+35=42n

Solution for 7n^2+35=42n equation:

7n^2+35=42n

We move all terms to the left:

7n^2+35-(42n)=0

a = 7; b = -42; c = +35;
Δ = b2-4ac
Δ = -422-4·7·35
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-28}{2*7}=\frac{14}{14}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+28}{2*7}=\frac{70}{14}
=5
$