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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3z^2+9z-4z+12)=0

We get rid of parentheses

-3z^2+9z-4z+12=0

We add all the numbers together, and all the variables

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

a = -3; b = 5; c = +12;
Δ = b2-4ac
Δ = 52-4·(-3)·12
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{-(5)-13}{2*-3}=\frac{-18}{-6}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*-3}=\frac{8}{-6}
=-1+1/3
$