7y(3y+6)=38

Solution for 7y(3y+6)=38 equation:

7y(3y+6)=38

We move all terms to the left:

7y(3y+6)-(38)=0

We multiply parentheses

21y^2+42y-38=0

a = 21; b = 42; c = -38;
Δ = b2-4ac
Δ = 422-4·21·(-38)
Δ = 4956
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4956}=\sqrt{4*1239}=\sqrt{4}*\sqrt{1239}=2\sqrt{1239}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{1239}}{2*21}=\frac{-42-2\sqrt{1239}}{42}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{1239}}{2*21}=\frac{-42+2\sqrt{1239}}{42}
$