(n+5)(n-4)=35

Solution for (n+5)(n-4)=35 equation:

(n+5)(n-4)=35

We move all terms to the left:

(n+5)(n-4)-(35)=0

We multiply parentheses ..

(+n^2-4n+5n-20)-35=0

We get rid of parentheses

n^2-4n+5n-20-35=0

We add all the numbers together, and all the variables

n^2+n-55=0

a = 1; b = 1; c = -55;
Δ = b2-4ac
Δ = 12-4·1·(-55)
Δ = 221
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{221}}{2*1}=\frac{-1-\sqrt{221}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{221}}{2*1}=\frac{-1+\sqrt{221}}{2}
$