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2b^2-9b-26=0
a = 2; b = -9; c = -26;
Δ = b2-4ac
Δ = -92-4·2·(-26)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-17}{2*2}=\frac{-8}{4} =-2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+17}{2*2}=\frac{26}{4} =6+1/2 $
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