(3z+7)(9-z)=0

Solution for (3z+7)(9-z)=0 equation:

(3z+7)(9-z)=0

We add all the numbers together, and all the variables

(3z+7)(-1z+9)=0

We multiply parentheses ..

(-3z^2+27z-7z+63)=0

We get rid of parentheses

-3z^2+27z-7z+63=0

We add all the numbers together, and all the variables

-3z^2+20z+63=0

a = -3; b = 20; c = +63;
Δ = b2-4ac
Δ = 202-4·(-3)·63
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{-(20)-34}{2*-3}=\frac{-54}{-6}
=+9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+34}{2*-3}=\frac{14}{-6}
=-2+1/3
$