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3x^2+2x-33=0
a = 3; b = 2; c = -33;
Δ = b2-4ac
Δ = 22-4·3·(-33)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-20}{2*3}=\frac{-22}{6} =-3+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+20}{2*3}=\frac{18}{6} =3 $
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