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3x^2-23x-120=0
a = 3; b = -23; c = -120;
Δ = b2-4ac
Δ = -232-4·3·(-120)
Δ = 1969
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{-(-23)-\sqrt{1969}}{2*3}=\frac{23-\sqrt{1969}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{1969}}{2*3}=\frac{23+\sqrt{1969}}{6} $
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