24z+3=2/4z+5

Solution for 24z+3=2/4z+5 equation:

24z+3=2/4z+5

We move all terms to the left:

24z+3-(2/4z+5)=0


Domain of the equation: 4z+5)!=0

z∈R


We get rid of parentheses

24z-2/4z-5+3=0

We multiply all the terms by the denominator


24z*4z-5*4z+3*4z-2=0

Wy multiply elements

96z^2-20z+12z-2=0

We add all the numbers together, and all the variables

96z^2-8z-2=0

a = 96; b = -8; c = -2;
Δ = b2-4ac
Δ = -82-4·96·(-2)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{13}}{2*96}=\frac{8-8\sqrt{13}}{192}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{13}}{2*96}=\frac{8+8\sqrt{13}}{192}
$