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3x^2+4x-165=0
a = 3; b = 4; c = -165;
Δ = b2-4ac
Δ = 42-4·3·(-165)
Δ = 1996
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1996}=\sqrt{4*499}=\sqrt{4}*\sqrt{499}=2\sqrt{499}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{499}}{2*3}=\frac{-4-2\sqrt{499}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{499}}{2*3}=\frac{-4+2\sqrt{499}}{6} $
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