25z^2+99=100z

Solution for 25z^2+99=100z equation:

25z^2+99=100z

We move all terms to the left:

25z^2+99-(100z)=0

a = 25; b = -100; c = +99;
Δ = b2-4ac
Δ = -1002-4·25·99
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-10}{2*25}=\frac{90}{50}
=1+4/5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+10}{2*25}=\frac{110}{50}
=2+1/5
$