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3x^2+22x-93=0
a = 3; b = 22; c = -93;
Δ = b2-4ac
Δ = 222-4·3·(-93)
Δ = 1600
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{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-40}{2*3}=\frac{-62}{6} =-10+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+40}{2*3}=\frac{18}{6} =3 $
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