(4z+1)(3-z)=0

Solution for (4z+1)(3-z)=0 equation:

(4z+1)(3-z)=0

We add all the numbers together, and all the variables

(4z+1)(-1z+3)=0

We multiply parentheses ..

(-4z^2+12z-1z+3)=0

We get rid of parentheses

-4z^2+12z-1z+3=0

We add all the numbers together, and all the variables

-4z^2+11z+3=0

a = -4; b = 11; c = +3;
Δ = b2-4ac
Δ = 112-4·(-4)·3
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*-4}=\frac{-24}{-8}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*-4}=\frac{2}{-8}
=-1/4
$