Y=(3p+4)(p+7)

Solution for Y=(3p+4)(p+7) equation:

=(3Y+4)(Y+7)

We move all terms to the left:

-((3Y+4)(Y+7))=0

We multiply parentheses ..

-((+3Y^2+21Y+4Y+28))=0


We calculate terms in parentheses: -((+3Y^2+21Y+4Y+28)), so:

(+3Y^2+21Y+4Y+28)

We get rid of parentheses

3Y^2+21Y+4Y+28

We add all the numbers together, and all the variables

3Y^2+25Y+28

Back to the equation:

-(3Y^2+25Y+28)


We get rid of parentheses

-3Y^2-25Y-28=0

a = -3; b = -25; c = -28;
Δ = b2-4ac
Δ = -252-4·(-3)·(-28)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-17}{2*-3}=\frac{8}{-6}
=-1+1/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+17}{2*-3}=\frac{42}{-6}
=-7
$