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6x^2-4x+(x+2)=11
We move all terms to the left:
6x^2-4x+(x+2)-(11)=0
We get rid of parentheses
6x^2-4x+x+2-11=0
We add all the numbers together, and all the variables
6x^2-3x-9=0
a = 6; b = -3; c = -9;
Δ = b2-4ac
Δ = -32-4·6·(-9)
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*6}=\frac{-12}{12} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*6}=\frac{18}{12} =1+1/2 $
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