3z(2z+11)=63

Solution for 3z(2z+11)=63 equation:

3z(2z+11)=63

We move all terms to the left:

3z(2z+11)-(63)=0

We multiply parentheses

6z^2+33z-63=0

a = 6; b = 33; c = -63;
Δ = b2-4ac
Δ = 332-4·6·(-63)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-51}{2*6}=\frac{-84}{12}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+51}{2*6}=\frac{18}{12}
=1+1/2
$