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27x^2-48x+19=0
a = 27; b = -48; c = +19;
Δ = b2-4ac
Δ = -482-4·27·19
Δ = 252
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-6\sqrt{7}}{2*27}=\frac{48-6\sqrt{7}}{54} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+6\sqrt{7}}{2*27}=\frac{48+6\sqrt{7}}{54} $
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