(z+3)(z+2)=42

Solution for (z+3)(z+2)=42 equation:

(z+3)(z+2)=42

We move all terms to the left:

(z+3)(z+2)-(42)=0

We multiply parentheses ..

(+z^2+2z+3z+6)-42=0

We get rid of parentheses

z^2+2z+3z+6-42=0

We add all the numbers together, and all the variables

z^2+5z-36=0

a = 1; b = 5; c = -36;
Δ = b2-4ac
Δ = 52-4·1·(-36)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*1}=\frac{-18}{2}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*1}=\frac{8}{2}
=4
$