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b^2-3b-42=0
a = 1; b = -3; c = -42;
Δ = b2-4ac
Δ = -32-4·1·(-42)
Δ = 177
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{177}}{2*1}=\frac{3-\sqrt{177}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{177}}{2*1}=\frac{3+\sqrt{177}}{2} $
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