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6x^2+33x-5=0
a = 6; b = 33; c = -5;
Δ = b2-4ac
Δ = 332-4·6·(-5)
Δ = 1209
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{1209}}{2*6}=\frac{-33-\sqrt{1209}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{1209}}{2*6}=\frac{-33+\sqrt{1209}}{12} $
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