(-21/2)z=-5

Solution for (-21/2)z=-5 equation:

(-21/2)z=-5

We move all terms to the left:

(-21/2)z-(-5)=0


Domain of the equation: 2)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(-21/2)z+5=0

We multiply parentheses

-21z^2+5=0

a = -21; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-21)·5
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{105}}{2*-21}=\frac{0-2\sqrt{105}}{-42}
=-\frac{2\sqrt{105}}{-42}
=-\frac{\sqrt{105}}{-21}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{105}}{2*-21}=\frac{0+2\sqrt{105}}{-42}
=\frac{2\sqrt{105}}{-42}
=\frac{\sqrt{105}}{-21}
$