(3y+1)(4y+3)=106

Solution for (3y+1)(4y+3)=106 equation:

(3y+1)(4y+3)=106

We move all terms to the left:

(3y+1)(4y+3)-(106)=0

We multiply parentheses ..

(+12y^2+9y+4y+3)-106=0

We get rid of parentheses

12y^2+9y+4y+3-106=0

We add all the numbers together, and all the variables

12y^2+13y-103=0

a = 12; b = 13; c = -103;
Δ = b2-4ac
Δ = 132-4·12·(-103)
Δ = 5113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{5113}}{2*12}=\frac{-13-\sqrt{5113}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{5113}}{2*12}=\frac{-13+\sqrt{5113}}{24}
$