3z(z-4)=5z-11

Solution for 3z(z-4)=5z-11 equation:

3z(z-4)=5z-11

We move all terms to the left:

3z(z-4)-(5z-11)=0

We multiply parentheses

3z^2-12z-(5z-11)=0

We get rid of parentheses

3z^2-12z-5z+11=0

We add all the numbers together, and all the variables

3z^2-17z+11=0

a = 3; b = -17; c = +11;
Δ = b2-4ac
Δ = -172-4·3·11
Δ = 157
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{157}}{2*3}=\frac{17-\sqrt{157}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{157}}{2*3}=\frac{17+\sqrt{157}}{6}
$