(3s-4)(2s-3)=0

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

(3s-4)(2s-3)=0

We multiply parentheses ..

(+6s^2-9s-8s+12)=0

We get rid of parentheses

6s^2-9s-8s+12=0

We add all the numbers together, and all the variables

6s^2-17s+12=0

a = 6; b = -17; c = +12;
Δ = b2-4ac
Δ = -172-4·6·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-1}{2*6}=\frac{16}{12}
=1+1/3
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+1}{2*6}=\frac{18}{12}
=1+1/2
$