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3x^2-35x-28=0
a = 3; b = -35; c = -28;
Δ = b2-4ac
Δ = -352-4·3·(-28)
Δ = 1561
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{-(-35)-\sqrt{1561}}{2*3}=\frac{35-\sqrt{1561}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1561}}{2*3}=\frac{35+\sqrt{1561}}{6} $
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