3x^2-26x+10=0

Solution for 3x^2-26x+10=0 equation:

3x^2-26x+10=0

a = 3; b = -26; c = +10;
Δ = b2-4ac
Δ = -262-4·3·10
Δ = 556
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:

$\sqrt{\Delta}=\sqrt{556}=\sqrt{4*139}=\sqrt{4}*\sqrt{139}=2\sqrt{139}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{139}}{2*3}=\frac{26-2\sqrt{139}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{139}}{2*3}=\frac{26+2\sqrt{139}}{6}
$