(a+4)(2a-3)=6

Solution for (a+4)(2a-3)=6 equation:

(a+4)(2a-3)=6

We move all terms to the left:

(a+4)(2a-3)-(6)=0

We multiply parentheses ..

(+2a^2-3a+8a-12)-6=0

We get rid of parentheses

2a^2-3a+8a-12-6=0

We add all the numbers together, and all the variables

2a^2+5a-18=0

a = 2; b = 5; c = -18;
Δ = b2-4ac
Δ = 52-4·2·(-18)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*2}=\frac{-18}{4}
=-4+1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*2}=\frac{8}{4}
=2
$