4(n-3)(n+2)=5

Solution for 4(n-3)(n+2)=5 equation:

4(n-3)(n+2)=5

We move all terms to the left:

4(n-3)(n+2)-(5)=0

We multiply parentheses ..

4(+n^2+2n-3n-6)-5=0

We multiply parentheses

4n^2+8n-12n-24-5=0

We add all the numbers together, and all the variables

4n^2-4n-29=0

a = 4; b = -4; c = -29;
Δ = b2-4ac
Δ = -42-4·4·(-29)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{30}}{2*4}=\frac{4-4\sqrt{30}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{30}}{2*4}=\frac{4+4\sqrt{30}}{8}
$