z(2z-1)+6=4(z+1)

Solution for z(2z-1)+6=4(z+1) equation:

z(2z-1)+6=4(z+1)

We move all terms to the left:

z(2z-1)+6-(4(z+1))=0

We multiply parentheses

2z^2-1z-(4(z+1))+6=0


We calculate terms in parentheses: -(4(z+1)), so:

4(z+1)

We multiply parentheses

4z+4

Back to the equation:

-(4z+4)


We get rid of parentheses

2z^2-1z-4z-4+6=0

We add all the numbers together, and all the variables

2z^2-5z+2=0

a = 2; b = -5; c = +2;
Δ = b2-4ac
Δ = -52-4·2·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2*2}=\frac{2}{4}
=1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3}{2*2}=\frac{8}{4}
=2
$