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

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

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

We multiply parentheses ..

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

We get rid of parentheses

6a^2-15a-8a+20=0

We add all the numbers together, and all the variables

6a^2-23a+20=0

a = 6; b = -23; c = +20;
Δ = b2-4ac
Δ = -232-4·6·20
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*6}=\frac{16}{12}
=1+1/3
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*6}=\frac{30}{12}
=2+1/2
$