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3x^2+2.6x-169=0
a = 3; b = 2.6; c = -169;
Δ = b2-4ac
Δ = 2.62-4·3·(-169)
Δ = 2034.76
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{-(2.6)-\sqrt{2034.76}}{2*3}=\frac{-2.6-\sqrt{2034.76}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.6)+\sqrt{2034.76}}{2*3}=\frac{-2.6+\sqrt{2034.76}}{6} $
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