x^2-26x+43=0

Solution for x^2-26x+43=0 equation:

x^2-26x+43=0

a = 1; b = -26; c = +43;
Δ = b2-4ac
Δ = -262-4·1·43
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6\sqrt{14}}{2*1}=\frac{26-6\sqrt{14}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6\sqrt{14}}{2*1}=\frac{26+6\sqrt{14}}{2}
$