Y=4(5x+3)(2x-1)

Solution for Y=4(5x+3)(2x-1) equation:

=4(5Y+3)(2Y-1)

We move all terms to the left:

-(4(5Y+3)(2Y-1))=0

We multiply parentheses ..

-(4(+10Y^2-5Y+6Y-3))=0


We calculate terms in parentheses: -(4(+10Y^2-5Y+6Y-3)), so:

4(+10Y^2-5Y+6Y-3)

We multiply parentheses

40Y^2-20Y+24Y-12

We add all the numbers together, and all the variables

40Y^2+4Y-12

Back to the equation:

-(40Y^2+4Y-12)


We get rid of parentheses

-40Y^2-4Y+12=0

a = -40; b = -4; c = +12;
Δ = b2-4ac
Δ = -42-4·(-40)·12
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-44}{2*-40}=\frac{-40}{-80}
=1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+44}{2*-40}=\frac{48}{-80}
=-3/5
$