25z^2-140=29

Solution for 25z^2-140=29 equation:

25z^2-140=29

We move all terms to the left:

25z^2-140-(29)=0

We add all the numbers together, and all the variables

25z^2-169=0

a = 25; b = 0; c = -169;
Δ = b2-4ac
Δ = 02-4·25·(-169)
Δ = 16900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16900}=130$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-130}{2*25}=\frac{-130}{50}
=-2+3/5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+130}{2*25}=\frac{130}{50}
=2+3/5
$