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l^2-31l+228=0
a = 1; b = -31; c = +228;
Δ = b2-4ac
Δ = -312-4·1·228
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-7}{2*1}=\frac{24}{2} =12 $$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+7}{2*1}=\frac{38}{2} =19 $
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