(n+3)(n-7)=48

Solution for (n+3)(n-7)=48 equation:

(n+3)(n-7)=48

We move all terms to the left:

(n+3)(n-7)-(48)=0

We multiply parentheses ..

(+n^2-7n+3n-21)-48=0

We get rid of parentheses

n^2-7n+3n-21-48=0

We add all the numbers together, and all the variables

n^2-4n-69=0

a = 1; b = -4; c = -69;
Δ = b2-4ac
Δ = -42-4·1·(-69)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{73}}{2*1}=\frac{4-2\sqrt{73}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{73}}{2*1}=\frac{4+2\sqrt{73}}{2}
$