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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

20k^2-7k-6=0

a = 20; b = -7; c = -6;
Δ = b2-4ac
Δ = -72-4·20·(-6)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-23}{2*20}=\frac{-16}{40}
=-2/5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+23}{2*20}=\frac{30}{40}
=3/4
$