3z(z-3)=7z+2

Solution for 3z(z-3)=7z+2 equation:

3z(z-3)=7z+2

We move all terms to the left:

3z(z-3)-(7z+2)=0

We multiply parentheses

3z^2-9z-(7z+2)=0

We get rid of parentheses

3z^2-9z-7z-2=0

We add all the numbers together, and all the variables

3z^2-16z-2=0

a = 3; b = -16; c = -2;
Δ = b2-4ac
Δ = -162-4·3·(-2)
Δ = 280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{70}}{2*3}=\frac{16-2\sqrt{70}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{70}}{2*3}=\frac{16+2\sqrt{70}}{6}
$