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2x^2+13x-29=0
a = 2; b = 13; c = -29;
Δ = b2-4ac
Δ = 132-4·2·(-29)
Δ = 401
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{-(13)-\sqrt{401}}{2*2}=\frac{-13-\sqrt{401}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{401}}{2*2}=\frac{-13+\sqrt{401}}{4} $
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