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3x^2-3x-40=0
a = 3; b = -3; c = -40;
Δ = b2-4ac
Δ = -32-4·3·(-40)
Δ = 489
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{-(-3)-\sqrt{489}}{2*3}=\frac{3-\sqrt{489}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{489}}{2*3}=\frac{3+\sqrt{489}}{6} $
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