(6y+14)(3y-4)=0

Solution for (6y+14)(3y-4)=0 equation:

(6y+14)(3y-4)=0

We multiply parentheses ..

(+18y^2-24y+42y-56)=0

We get rid of parentheses

18y^2-24y+42y-56=0

We add all the numbers together, and all the variables

18y^2+18y-56=0

a = 18; b = 18; c = -56;
Δ = b2-4ac
Δ = 182-4·18·(-56)
Δ = 4356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4356}=66$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-66}{2*18}=\frac{-84}{36}
=-2+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+66}{2*18}=\frac{48}{36}
=1+1/3
$