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3x(x+6)+x=149
We move all terms to the left:
3x(x+6)+x-(149)=0
We add all the numbers together, and all the variables
x+3x(x+6)-149=0
We multiply parentheses
3x^2+x+18x-149=0
We add all the numbers together, and all the variables
3x^2+19x-149=0
a = 3; b = 19; c = -149;
Δ = b2-4ac
Δ = 192-4·3·(-149)
Δ = 2149
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{2149}}{2*3}=\frac{-19-\sqrt{2149}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{2149}}{2*3}=\frac{-19+\sqrt{2149}}{6} $
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