Y=(4x+3)(2x+7)

Solution for Y=(4x+3)(2x+7) equation:

=(4Y+3)(2Y+7)

We move all terms to the left:

-((4Y+3)(2Y+7))=0

We multiply parentheses ..

-((+8Y^2+28Y+6Y+21))=0


We calculate terms in parentheses: -((+8Y^2+28Y+6Y+21)), so:

(+8Y^2+28Y+6Y+21)

We get rid of parentheses

8Y^2+28Y+6Y+21

We add all the numbers together, and all the variables

8Y^2+34Y+21

Back to the equation:

-(8Y^2+34Y+21)


We get rid of parentheses

-8Y^2-34Y-21=0

a = -8; b = -34; c = -21;
Δ = b2-4ac
Δ = -342-4·(-8)·(-21)
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{-(-34)-22}{2*-8}=\frac{12}{-16}
=-3/4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+22}{2*-8}=\frac{56}{-16}
=-3+1/2
$