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(2x-1)(3x+2)=3(x+2)
We move all terms to the left:
(2x-1)(3x+2)-(3(x+2))=0
We multiply parentheses ..
(+6x^2+4x-3x-2)-(3(x+2))=0
We calculate terms in parentheses: -(3(x+2)), so:We get rid of parentheses
3(x+2)
We multiply parentheses
3x+6
Back to the equation:
-(3x+6)
6x^2+4x-3x-3x-2-6=0
We add all the numbers together, and all the variables
6x^2-2x-8=0
a = 6; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·6·(-8)
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*6}=\frac{-12}{12} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*6}=\frac{16}{12} =1+1/3 $
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