1/4z+1=10(3z)+z

Solution for 1/4z+1=10(3z)+z equation:

1/4z+1=10(3z)+z

We move all terms to the left:

1/4z+1-(10(3z)+z)=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-(+104z)+1=0

We get rid of parentheses

1/4z-104z+1=0

We multiply all the terms by the denominator


-104z*4z+1*4z+1=0

Wy multiply elements

-416z^2+4z+1=0

a = -416; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·(-416)·1
Δ = 1680
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:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{105}}{2*-416}=\frac{-4-4\sqrt{105}}{-832}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{105}}{2*-416}=\frac{-4+4\sqrt{105}}{-832}
$