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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

(+2z^2+4z-1z-2)-2z-1=0

We get rid of parentheses

2z^2+4z-1z-2z-2-1=0

We add all the numbers together, and all the variables

2z^2+z-3=0

a = 2; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·2·(-3)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*2}=\frac{-6}{4}
=-1+1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*2}=\frac{4}{4}
=1
$