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7b^2-3b-17=0
a = 7; b = -3; c = -17;
Δ = b2-4ac
Δ = -32-4·7·(-17)
Δ = 485
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{485}}{2*7}=\frac{3-\sqrt{485}}{14} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{485}}{2*7}=\frac{3+\sqrt{485}}{14} $
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