Y=(5t+8)(2t-3)

Solution for Y=(5t+8)(2t-3) equation:

=(5Y+8)(2Y-3)

We move all terms to the left:

-((5Y+8)(2Y-3))=0

We multiply parentheses ..

-((+10Y^2-15Y+16Y-24))=0


We calculate terms in parentheses: -((+10Y^2-15Y+16Y-24)), so:

(+10Y^2-15Y+16Y-24)

We get rid of parentheses

10Y^2-15Y+16Y-24

We add all the numbers together, and all the variables

10Y^2+Y-24

Back to the equation:

-(10Y^2+Y-24)


We get rid of parentheses

-10Y^2-Y+24=0

We add all the numbers together, and all the variables

-10Y^2-1Y+24=0

a = -10; b = -1; c = +24;
Δ = b2-4ac
Δ = -12-4·(-10)·24
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-31}{2*-10}=\frac{-30}{-20}
=1+1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+31}{2*-10}=\frac{32}{-20}
=-1+3/5
$