y^2-25y+10=0

Solution for y^2-25y+10=0 equation:

y^2-25y+10=0

a = 1; b = -25; c = +10;
Δ = b2-4ac
Δ = -252-4·1·10
Δ = 585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-3\sqrt{65}}{2*1}=\frac{25-3\sqrt{65}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+3\sqrt{65}}{2*1}=\frac{25+3\sqrt{65}}{2}
$