1/4z-5=8z-36

Solution for 1/4z-5=8z-36 equation:

1/4z-5=8z-36

We move all terms to the left:

1/4z-5-(8z-36)=0


Domain of the equation: 4z!=0

z!=0/4

z!=0

z∈R


We get rid of parentheses

1/4z-8z+36-5=0

We multiply all the terms by the denominator


-8z*4z+36*4z-5*4z+1=0

Wy multiply elements

-32z^2+144z-20z+1=0

We add all the numbers together, and all the variables

-32z^2+124z+1=0

a = -32; b = 124; c = +1;
Δ = b2-4ac
Δ = 1242-4·(-32)·1
Δ = 15504
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:

$\sqrt{\Delta}=\sqrt{15504}=\sqrt{16*969}=\sqrt{16}*\sqrt{969}=4\sqrt{969}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(124)-4\sqrt{969}}{2*-32}=\frac{-124-4\sqrt{969}}{-64}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(124)+4\sqrt{969}}{2*-32}=\frac{-124+4\sqrt{969}}{-64}
$