3z(3z+19)=42

Solution for 3z(3z+19)=42 equation:

3z(3z+19)=42

We move all terms to the left:

3z(3z+19)-(42)=0

We multiply parentheses

9z^2+57z-42=0

a = 9; b = 57; c = -42;
Δ = b2-4ac
Δ = 572-4·9·(-42)
Δ = 4761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4761}=69$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-69}{2*9}=\frac{-126}{18}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+69}{2*9}=\frac{12}{18}
=2/3
$