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3b^2-9b-120=0
a = 3; b = -9; c = -120;
Δ = b2-4ac
Δ = -92-4·3·(-120)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1521}=39$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-39}{2*3}=\frac{-30}{6} =-5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+39}{2*3}=\frac{48}{6} =8 $
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