25z^2-100z+98=0

Solution for 25z^2-100z+98=0 equation:

25z^2-100z+98=0

a = 25; b = -100; c = +98;
Δ = b2-4ac
Δ = -1002-4·25·98
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-10\sqrt{2}}{2*25}=\frac{100-10\sqrt{2}}{50}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+10\sqrt{2}}{2*25}=\frac{100+10\sqrt{2}}{50}
$