3x2-15x+6=0

Solution for 3x2-15x+6=0 equation:

3x^2-15x+6=0

a = 3; b = -15; c = +6;
Δ = b2-4ac
Δ = -152-4·3·6
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{17}}{2*3}=\frac{15-3\sqrt{17}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{17}}{2*3}=\frac{15+3\sqrt{17}}{6}
$