(7z+2)(3z-4)=0

Solution for (7z+2)(3z-4)=0 equation:

(7z+2)(3z-4)=0

We multiply parentheses ..

(+21z^2-28z+6z-8)=0

We get rid of parentheses

21z^2-28z+6z-8=0

We add all the numbers together, and all the variables

21z^2-22z-8=0

a = 21; b = -22; c = -8;
Δ = b2-4ac
Δ = -222-4·21·(-8)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-34}{2*21}=\frac{-12}{42}
=-2/7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+34}{2*21}=\frac{56}{42}
=1+1/3
$