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

Solution for 0=(5y+3)(8y-3) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+40y^2-15y+24y-9))=0


We calculate terms in parentheses: -((+40y^2-15y+24y-9)), so:

(+40y^2-15y+24y-9)

We get rid of parentheses

40y^2-15y+24y-9

We add all the numbers together, and all the variables

40y^2+9y-9

Back to the equation:

-(40y^2+9y-9)


We get rid of parentheses

-40y^2-9y+9=0

a = -40; b = -9; c = +9;
Δ = b2-4ac
Δ = -92-4·(-40)·9
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-39}{2*-40}=\frac{-30}{-80}
=3/8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+39}{2*-40}=\frac{48}{-80}
=-3/5
$