3x2-27x+48=0

Solution for 3x2-27x+48=0 equation:

3x^2-27x+48=0

a = 3; b = -27; c = +48;
Δ = b2-4ac
Δ = -272-4·3·48
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{-(-27)-3\sqrt{17}}{2*3}=\frac{27-3\sqrt{17}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3\sqrt{17}}{2*3}=\frac{27+3\sqrt{17}}{6}
$