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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+8Y^2-10Y+12Y-15))=0


We calculate terms in parentheses: -((+8Y^2-10Y+12Y-15)), so:

(+8Y^2-10Y+12Y-15)

We get rid of parentheses

8Y^2-10Y+12Y-15

We add all the numbers together, and all the variables

8Y^2+2Y-15

Back to the equation:

-(8Y^2+2Y-15)


We get rid of parentheses

-8Y^2-2Y+15=0

a = -8; b = -2; c = +15;
Δ = b2-4ac
Δ = -22-4·(-8)·15
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*-8}=\frac{-20}{-16}
=1+1/4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*-8}=\frac{24}{-16}
=-1+1/2
$