(2z+5)(z+1)=-1

Solution for (2z+5)(z+1)=-1 equation:

(2z+5)(z+1)=-1

We move all terms to the left:

(2z+5)(z+1)-(-1)=0

We add all the numbers together, and all the variables

(2z+5)(z+1)+1=0

We multiply parentheses ..

(+2z^2+2z+5z+5)+1=0

We get rid of parentheses

2z^2+2z+5z+5+1=0

We add all the numbers together, and all the variables

2z^2+7z+6=0

a = 2; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·2·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*2}=\frac{-8}{4}
=-2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*2}=\frac{-6}{4}
=-1+1/2
$