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

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

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

We multiply parentheses ..

(+10y^2-15y+8y-12)=0

We get rid of parentheses

10y^2-15y+8y-12=0

We add all the numbers together, and all the variables

10y^2-7y-12=0

a = 10; b = -7; c = -12;
Δ = b2-4ac
Δ = -72-4·10·(-12)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-23}{2*10}=\frac{-16}{20}
=-4/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+23}{2*10}=\frac{30}{20}
=1+1/2
$