(z+3)(z+2)=z+3z+20

Solution for (z+3)(z+2)=z+3z+20 equation:

(z+3)(z+2)=z+3z+20

We move all terms to the left:

(z+3)(z+2)-(z+3z+20)=0

We add all the numbers together, and all the variables

(z+3)(z+2)-(4z+20)=0

We get rid of parentheses

(z+3)(z+2)-4z-20=0

We multiply parentheses ..

(+z^2+2z+3z+6)-4z-20=0

We get rid of parentheses

z^2+2z+3z-4z+6-20=0

We add all the numbers together, and all the variables

z^2+z-14=0

a = 1; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·1·(-14)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{57}}{2*1}=\frac{-1-\sqrt{57}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{57}}{2*1}=\frac{-1+\sqrt{57}}{2}
$