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8x^2+x-9=0
a = 8; b = 1; c = -9;
Δ = b2-4ac
Δ = 12-4·8·(-9)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*8}=\frac{-18}{16} =-1+1/8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*8}=\frac{16}{16} =1 $
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