X^2-26x+42=0

Solution for X^2-26x+42=0 equation:

X^2-26X+42=0

a = 1; b = -26; c = +42;
Δ = b2-4ac
Δ = -262-4·1·42
Δ = 508
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:

$\sqrt{\Delta}=\sqrt{508}=\sqrt{4*127}=\sqrt{4}*\sqrt{127}=2\sqrt{127}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{127}}{2*1}=\frac{26-2\sqrt{127}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{127}}{2*1}=\frac{26+2\sqrt{127}}{2}
$