3z(3z+14)=15

Solution for 3z(3z+14)=15 equation:

3z(3z+14)=15

We move all terms to the left:

3z(3z+14)-(15)=0

We multiply parentheses

9z^2+42z-15=0

a = 9; b = 42; c = -15;
Δ = b2-4ac
Δ = 422-4·9·(-15)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-48}{2*9}=\frac{-90}{18}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+48}{2*9}=\frac{6}{18}
=1/3
$