(2/5)z+4=13

Solution for (2/5)z+4=13 equation:

(2/5)z+4=13

We move all terms to the left:

(2/5)z+4-(13)=0


Domain of the equation: 5)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+2/5)z+4-13=0

We add all the numbers together, and all the variables

(+2/5)z-9=0

We multiply parentheses

2z^2-9=0

a = 2; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·2·(-9)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*2}=\frac{0-6\sqrt{2}}{4}
=-\frac{6\sqrt{2}}{4}
=-\frac{3\sqrt{2}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*2}=\frac{0+6\sqrt{2}}{4}
=\frac{6\sqrt{2}}{4}
=\frac{3\sqrt{2}}{2}
$