(2n-7)(n+2)=6

Solution for (2n-7)(n+2)=6 equation:

(2n-7)(n+2)=6

We move all terms to the left:

(2n-7)(n+2)-(6)=0

We multiply parentheses ..

(+2n^2+4n-7n-14)-6=0

We get rid of parentheses

2n^2+4n-7n-14-6=0

We add all the numbers together, and all the variables

2n^2-3n-20=0

a = 2; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·2·(-20)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*2}=\frac{-10}{4}
=-2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*2}=\frac{16}{4}
=4
$