(3z-5)(6+z)=0

Solution for (3z-5)(6+z)=0 equation:

(3z-5)(6+z)=0

We add all the numbers together, and all the variables

(3z-5)(z+6)=0

We multiply parentheses ..

(+3z^2+18z-5z-30)=0

We get rid of parentheses

3z^2+18z-5z-30=0

We add all the numbers together, and all the variables

3z^2+13z-30=0

a = 3; b = 13; c = -30;
Δ = b2-4ac
Δ = 132-4·3·(-30)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*3}=\frac{-36}{6}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*3}=\frac{10}{6}
=1+2/3
$