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f^2+23f+22=0
a = 1; b = 23; c = +22;
Δ = b2-4ac
Δ = 232-4·1·22
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-21}{2*1}=\frac{-44}{2} =-22 $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+21}{2*1}=\frac{-2}{2} =-1 $
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