3x^2-26x+47=0

Solution for 3x^2-26x+47=0 equation:

3x^2-26x+47=0

a = 3; b = -26; c = +47;
Δ = b2-4ac
Δ = -262-4·3·47
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-4\sqrt{7}}{2*3}=\frac{26-4\sqrt{7}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+4\sqrt{7}}{2*3}=\frac{26+4\sqrt{7}}{6}
$