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(2x+3)(3x+1)=11x+27
We move all terms to the left:
(2x+3)(3x+1)-(11x+27)=0
We get rid of parentheses
(2x+3)(3x+1)-11x-27=0
We multiply parentheses ..
(+6x^2+2x+9x+3)-11x-27=0
We get rid of parentheses
6x^2+2x+9x-11x+3-27=0
We add all the numbers together, and all the variables
6x^2-24=0
a = 6; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·6·(-24)
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*6}=\frac{-24}{12} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*6}=\frac{24}{12} =2 $
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