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6x^2+14x-40=0
a = 6; b = 14; c = -40;
Δ = b2-4ac
Δ = 142-4·6·(-40)
Δ = 1156
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{1156}=34$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-34}{2*6}=\frac{-48}{12} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+34}{2*6}=\frac{20}{12} =1+2/3 $
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