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3l^2-6l-105=0
a = 3; b = -6; c = -105;
Δ = b2-4ac
Δ = -62-4·3·(-105)
Δ = 1296
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{1296}=36$$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-36}{2*3}=\frac{-30}{6} =-5 $$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+36}{2*3}=\frac{42}{6} =7 $
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