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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-6x^2+23x-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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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