(5y-3)(2y+1)=A

Solution for (5y-3)(2y+1)=A equation:

(5y-3)(2y+1)=

We move all terms to the left:

(5y-3)(2y+1)-()=0

We add all the numbers together, and all the variables

(5y-3)(2y+1)=0

We multiply parentheses ..

(+10y^2+5y-6y-3)=0

We get rid of parentheses

10y^2+5y-6y-3=0

We add all the numbers together, and all the variables

10y^2-1y-3=0

a = 10; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·10·(-3)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*10}=\frac{-10}{20}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*10}=\frac{12}{20}
=3/5
$