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3x^2-49x+24=0
a = 3; b = -49; c = +24;
Δ = b2-4ac
Δ = -492-4·3·24
Δ = 2113
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{-(-49)-\sqrt{2113}}{2*3}=\frac{49-\sqrt{2113}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{2113}}{2*3}=\frac{49+\sqrt{2113}}{6} $
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