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-2x2+13x+20=0

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Solution for -2x2+13x+20=0 equation:



-2x^2+13x+20=0
a = -2; b = 13; c = +20;
Δ = b2-4ac
Δ = 132-4·(-2)·20
Δ = 329
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{329}}{2*-2}=\frac{-13-\sqrt{329}}{-4}
x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{329}}{2*-2}=\frac{-13+\sqrt{329}}{-4}

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