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8x^2+39x-5=0
a = 8; b = 39; c = -5;
Δ = b2-4ac
Δ = 392-4·8·(-5)
Δ = 1681
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{1681}=41$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-41}{2*8}=\frac{-80}{16} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+41}{2*8}=\frac{2}{16} =1/8 $
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