3+(1/4)z=31

Solution for 3+(1/4)z=31 equation:

3+(1/4)z=31

We move all terms to the left:

3+(1/4)z-(31)=0


Domain of the equation: 4)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+1/4)z+3-31=0

We add all the numbers together, and all the variables

(+1/4)z-28=0

We multiply parentheses

z^2-28=0

a = 1; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·1·(-28)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{7}}{2*1}=\frac{0-4\sqrt{7}}{2}
=-\frac{4\sqrt{7}}{2}
=-2\sqrt{7}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{7}}{2*1}=\frac{0+4\sqrt{7}}{2}
=\frac{4\sqrt{7}}{2}
=2\sqrt{7}
$