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b2+3b-22=0
We add all the numbers together, and all the variables
b^2+3b-22=0
a = 1; b = 3; c = -22;
Δ = b2-4ac
Δ = 32-4·1·(-22)
Δ = 97
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{97}}{2*1}=\frac{-3-\sqrt{97}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{97}}{2*1}=\frac{-3+\sqrt{97}}{2} $
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