25i^2=10i+1=0

Solution for 25i^2=10i+1=0 equation:

25i^2=10i+1=0

We move all terms to the left:

25i^2-(10i+1)=0

We get rid of parentheses

25i^2-10i-1=0

a = 25; b = -10; c = -1;
Δ = b2-4ac
Δ = -102-4·25·(-1)
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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