27x^2-48x-5=0

Solution for 27x^2-48x-5=0 equation:

27x^2-48x-5=0

a = 27; b = -48; c = -5;
Δ = b2-4ac
Δ = -482-4·27·(-5)
Δ = 2844
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:

$\sqrt{\Delta}=\sqrt{2844}=\sqrt{36*79}=\sqrt{36}*\sqrt{79}=6\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-6\sqrt{79}}{2*27}=\frac{48-6\sqrt{79}}{54}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+6\sqrt{79}}{2*27}=\frac{48+6\sqrt{79}}{54}
$