(3z+23z)z=5

Solution for (3z+23z)z=5 equation:

(3z+23z)z=5

We move all terms to the left:

(3z+23z)z-(5)=0

We add all the numbers together, and all the variables

(+26z)z-5=0

We multiply parentheses

26z^2-5=0

a = 26; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·26·(-5)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{130}}{2*26}=\frac{0-2\sqrt{130}}{52}
=-\frac{2\sqrt{130}}{52}
=-\frac{\sqrt{130}}{26}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{130}}{2*26}=\frac{0+2\sqrt{130}}{52}
=\frac{2\sqrt{130}}{52}
=\frac{\sqrt{130}}{26}
$