4z(z+3)=4z+5

Solution for 4z(z+3)=4z+5 equation:

4z(z+3)=4z+5

We move all terms to the left:

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

We multiply parentheses

4z^2+12z-(4z+5)=0

We get rid of parentheses

4z^2+12z-4z-5=0

We add all the numbers together, and all the variables

4z^2+8z-5=0

a = 4; b = 8; c = -5;
Δ = b2-4ac
Δ = 82-4·4·(-5)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*4}=\frac{-20}{8}
=-2+1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*4}=\frac{4}{8}
=1/2
$