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3x^2+33x-540=0
a = 3; b = 33; c = -540;
Δ = b2-4ac
Δ = 332-4·3·(-540)
Δ = 7569
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{7569}=87$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-87}{2*3}=\frac{-120}{6} =-20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+87}{2*3}=\frac{54}{6} =9 $
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