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5x^2+39x-28=0
a = 5; b = 39; c = -28;
Δ = b2-4ac
Δ = 392-4·5·(-28)
Δ = 2081
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{-(39)-\sqrt{2081}}{2*5}=\frac{-39-\sqrt{2081}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{2081}}{2*5}=\frac{-39+\sqrt{2081}}{10} $
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