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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+5Y^2-4Y+15Y-12))=0


We calculate terms in parentheses: -((+5Y^2-4Y+15Y-12)), so:

(+5Y^2-4Y+15Y-12)

We get rid of parentheses

5Y^2-4Y+15Y-12

We add all the numbers together, and all the variables

5Y^2+11Y-12

Back to the equation:

-(5Y^2+11Y-12)


We get rid of parentheses

-5Y^2-11Y+12=0

a = -5; b = -11; c = +12;
Δ = b2-4ac
Δ = -112-4·(-5)·12
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*-5}=\frac{-8}{-10}
=4/5
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*-5}=\frac{30}{-10}
=-3
$