5x^2-26x+14=0

Solution for 5x^2-26x+14=0 equation:

5x^2-26x+14=0

a = 5; b = -26; c = +14;
Δ = b2-4ac
Δ = -262-4·5·14
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6\sqrt{11}}{2*5}=\frac{26-6\sqrt{11}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6\sqrt{11}}{2*5}=\frac{26+6\sqrt{11}}{10}
$