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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

(5y-3)(y+3)-4y+8-3=0

We multiply parentheses ..

(+5y^2+15y-3y-9)-4y+8-3=0

We add all the numbers together, and all the variables

(+5y^2+15y-3y-9)-4y+5=0

We get rid of parentheses

5y^2+15y-3y-4y-9+5=0

We add all the numbers together, and all the variables

5y^2+8y-4=0

a = 5; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·5·(-4)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*5}=\frac{-20}{10}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*5}=\frac{4}{10}
=2/5
$