2-2z=3/4z+13

Solution for 2-2z=3/4z+13 equation:

2-2z=3/4z+13

We move all terms to the left:

2-2z-(3/4z+13)=0


Domain of the equation: 4z+13)!=0

z∈R


We get rid of parentheses

-2z-3/4z-13+2=0

We multiply all the terms by the denominator


-2z*4z-13*4z+2*4z-3=0

Wy multiply elements

-8z^2-52z+8z-3=0

We add all the numbers together, and all the variables

-8z^2-44z-3=0

a = -8; b = -44; c = -3;
Δ = b2-4ac
Δ = -442-4·(-8)·(-3)
Δ = 1840
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:

$\sqrt{\Delta}=\sqrt{1840}=\sqrt{16*115}=\sqrt{16}*\sqrt{115}=4\sqrt{115}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-4\sqrt{115}}{2*-8}=\frac{44-4\sqrt{115}}{-16}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+4\sqrt{115}}{2*-8}=\frac{44+4\sqrt{115}}{-16}
$