(n-3)(n+4)=10

Solution for (n-3)(n+4)=10 equation:

(n-3)(n+4)=10

We move all terms to the left:

(n-3)(n+4)-(10)=0

We multiply parentheses ..

(+n^2+4n-3n-12)-10=0

We get rid of parentheses

n^2+4n-3n-12-10=0

We add all the numbers together, and all the variables

n^2+n-22=0

a = 1; b = 1; c = -22;
Δ = b2-4ac
Δ = 12-4·1·(-22)
Δ = 89
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{89}}{2*1}=\frac{-1-\sqrt{89}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{89}}{2*1}=\frac{-1+\sqrt{89}}{2}
$