2z+4(z+5)=1/2z+5

Solution for 2z+4(z+5)=1/2z+5 equation:

2z+4(z+5)=1/2z+5

We move all terms to the left:

2z+4(z+5)-(1/2z+5)=0


Domain of the equation: 2z+5)!=0

z∈R


We multiply parentheses

2z+4z-(1/2z+5)+20=0

We get rid of parentheses

2z+4z-1/2z-5+20=0

We multiply all the terms by the denominator


2z*2z+4z*2z-5*2z+20*2z-1=0

Wy multiply elements

4z^2+8z^2-10z+40z-1=0

We add all the numbers together, and all the variables

12z^2+30z-1=0

a = 12; b = 30; c = -1;
Δ = b2-4ac
Δ = 302-4·12·(-1)
Δ = 948
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{948}=\sqrt{4*237}=\sqrt{4}*\sqrt{237}=2\sqrt{237}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{237}}{2*12}=\frac{-30-2\sqrt{237}}{24}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{237}}{2*12}=\frac{-30+2\sqrt{237}}{24}
$