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3x^2+6x-48=0
a = 3; b = 6; c = -48;
Δ = b2-4ac
Δ = 62-4·3·(-48)
Δ = 612
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{17}}{2*3}=\frac{-6-6\sqrt{17}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{17}}{2*3}=\frac{-6+6\sqrt{17}}{6} $
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