(2z+5)(6+z)=0

Solution for (2z+5)(6+z)=0 equation:

(2z+5)(6+z)=0

We add all the numbers together, and all the variables

(2z+5)(z+6)=0

We multiply parentheses ..

(+2z^2+12z+5z+30)=0

We get rid of parentheses

2z^2+12z+5z+30=0

We add all the numbers together, and all the variables

2z^2+17z+30=0

a = 2; b = 17; c = +30;
Δ = b2-4ac
Δ = 172-4·2·30
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*2}=\frac{-24}{4}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*2}=\frac{-10}{4}
=-2+1/2
$