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3x^2-33x-38=0
a = 3; b = -33; c = -38;
Δ = b2-4ac
Δ = -332-4·3·(-38)
Δ = 1545
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{-(-33)-\sqrt{1545}}{2*3}=\frac{33-\sqrt{1545}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{1545}}{2*3}=\frac{33+\sqrt{1545}}{6} $
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