1/4z-3=5-2z

Solution for 1/4z-3=5-2z equation:

1/4z-3=5-2z

We move all terms to the left:

1/4z-3-(5-2z)=0


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R


We add all the numbers together, and all the variables

1/4z-(-2z+5)-3=0

We get rid of parentheses

1/4z+2z-5-3=0

We multiply all the terms by the denominator


2z*4z-5*4z-3*4z+1=0

Wy multiply elements

8z^2-20z-12z+1=0

We add all the numbers together, and all the variables

8z^2-32z+1=0

a = 8; b = -32; c = +1;
Δ = b2-4ac
Δ = -322-4·8·1
Δ = 992
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:

$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{62}}{2*8}=\frac{32-4\sqrt{62}}{16}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{62}}{2*8}=\frac{32+4\sqrt{62}}{16}
$