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(3x)(x+2)=105
We move all terms to the left:
(3x)(x+2)-(105)=0
We multiply parentheses
3x^2+6x-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:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-36}{2*3}=\frac{-42}{6} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+36}{2*3}=\frac{30}{6} =5 $
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