2n^2+10=9n

Solution for 2n^2+10=9n equation:

2n^2+10=9n

We move all terms to the left:

2n^2+10-(9n)=0

a = 2; b = -9; c = +10;
Δ = b2-4ac
Δ = -92-4·2·10
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-1}{2*2}=\frac{8}{4}
=2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*2}=\frac{10}{4}
=2+1/2
$