2n^2+41n+1=10n+9

Solution for 2n^2+41n+1=10n+9 equation:

2n^2+41n+1=10n+9

We move all terms to the left:

2n^2+41n+1-(10n+9)=0

We get rid of parentheses

2n^2+41n-10n-9+1=0

We add all the numbers together, and all the variables

2n^2+31n-8=0

a = 2; b = 31; c = -8;
Δ = b2-4ac
Δ = 312-4·2·(-8)
Δ = 1025
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:

$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-5\sqrt{41}}{2*2}=\frac{-31-5\sqrt{41}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+5\sqrt{41}}{2*2}=\frac{-31+5\sqrt{41}}{4}
$