x^2-26x+35=0

Solution for x^2-26x+35=0 equation:

x^2-26x+35=0

a = 1; b = -26; c = +35;
Δ = b2-4ac
Δ = -262-4·1·35
Δ = 536
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{134}}{2*1}=\frac{26-2\sqrt{134}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{134}}{2*1}=\frac{26+2\sqrt{134}}{2}
$