y^2-26y+40=0

Solution for y^2-26y+40=0 equation:

y^2-26y+40=0

a = 1; b = -26; c = +40;
Δ = b2-4ac
Δ = -262-4·1·40
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{129}}{2*1}=\frac{26-2\sqrt{129}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{129}}{2*1}=\frac{26+2\sqrt{129}}{2}
$