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6x^2+25x-129=0
a = 6; b = 25; c = -129;
Δ = b2-4ac
Δ = 252-4·6·(-129)
Δ = 3721
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{3721}=61$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-61}{2*6}=\frac{-86}{12} =-7+1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+61}{2*6}=\frac{36}{12} =3 $
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