(3a-4)(2a+1)-(6a-9)=15

Solution for (3a-4)(2a+1)-(6a-9)=15 equation:

(3a-4)(2a+1)-(6a-9)=15

We move all terms to the left:

(3a-4)(2a+1)-(6a-9)-(15)=0

We get rid of parentheses

(3a-4)(2a+1)-6a+9-15=0

We multiply parentheses ..

(+6a^2+3a-8a-4)-6a+9-15=0

We add all the numbers together, and all the variables

(+6a^2+3a-8a-4)-6a-6=0

We get rid of parentheses

6a^2+3a-8a-6a-4-6=0

We add all the numbers together, and all the variables

6a^2-11a-10=0

a = 6; b = -11; c = -10;
Δ = b2-4ac
Δ = -112-4·6·(-10)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*6}=\frac{-8}{12}
=-2/3
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*6}=\frac{30}{12}
=2+1/2
$