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3x^2+69x-28=0
a = 3; b = 69; c = -28;
Δ = b2-4ac
Δ = 692-4·3·(-28)
Δ = 5097
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{-(69)-\sqrt{5097}}{2*3}=\frac{-69-\sqrt{5097}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(69)+\sqrt{5097}}{2*3}=\frac{-69+\sqrt{5097}}{6} $
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