4z+12-4=81/2z+8

Solution for 4z+12-4=81/2z+8 equation:

4z+12-4=81/2z+8

We move all terms to the left:

4z+12-4-(81/2z+8)=0


Domain of the equation: 2z+8)!=0

z∈R


We add all the numbers together, and all the variables

4z-(81/2z+8)+8=0

We get rid of parentheses

4z-81/2z-8+8=0

We multiply all the terms by the denominator


4z*2z-8*2z+8*2z-81=0

Wy multiply elements

8z^2-16z+16z-81=0

We add all the numbers together, and all the variables

8z^2-81=0

a = 8; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·8·(-81)
Δ = 2592
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:

$\sqrt{\Delta}=\sqrt{2592}=\sqrt{1296*2}=\sqrt{1296}*\sqrt{2}=36\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{2}}{2*8}=\frac{0-36\sqrt{2}}{16}
=-\frac{36\sqrt{2}}{16}
=-\frac{9\sqrt{2}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{2}}{2*8}=\frac{0+36\sqrt{2}}{16}
=\frac{36\sqrt{2}}{16}
=\frac{9\sqrt{2}}{4}
$