(2z-9)(6+z)=0

Solution for (2z-9)(6+z)=0 equation:

(2z-9)(6+z)=0

We add all the numbers together, and all the variables

(2z-9)(z+6)=0

We multiply parentheses ..

(+2z^2+12z-9z-54)=0

We get rid of parentheses

2z^2+12z-9z-54=0

We add all the numbers together, and all the variables

2z^2+3z-54=0

a = 2; b = 3; c = -54;
Δ = b2-4ac
Δ = 32-4·2·(-54)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*2}=\frac{-24}{4}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*2}=\frac{18}{4}
=4+1/2
$