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3x^2-119x-324=0
a = 3; b = -119; c = -324;
Δ = b2-4ac
Δ = -1192-4·3·(-324)
Δ = 18049
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-119)-\sqrt{18049}}{2*3}=\frac{119-\sqrt{18049}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-119)+\sqrt{18049}}{2*3}=\frac{119+\sqrt{18049}}{6} $
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