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

Solution for (5a-3)(a+4)-(2a+5)(3a-4)=0 equation:

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

We multiply parentheses ..

(+5a^2+20a-3a-12)-(2a+5)(3a-4)=0

We get rid of parentheses

5a^2+20a-3a-(2a+5)(3a-4)-12=0

We multiply parentheses ..

5a^2-(+6a^2-8a+15a-20)+20a-3a-12=0

We add all the numbers together, and all the variables

5a^2-(+6a^2-8a+15a-20)+17a-12=0

We get rid of parentheses

5a^2-6a^2+8a-15a+17a+20-12=0

We add all the numbers together, and all the variables

-1a^2+10a+8=0

a = -1; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·(-1)·8
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{33}}{2*-1}=\frac{-10-2\sqrt{33}}{-2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{33}}{2*-1}=\frac{-10+2\sqrt{33}}{-2}
$