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3x(x+2)-20=5(x-2)
We move all terms to the left:
3x(x+2)-20-(5(x-2))=0
We multiply parentheses
3x^2+6x-(5(x-2))-20=0
We calculate terms in parentheses: -(5(x-2)), so:We get rid of parentheses
5(x-2)
We multiply parentheses
5x-10
Back to the equation:
-(5x-10)
3x^2+6x-5x+10-20=0
We add all the numbers together, and all the variables
3x^2+x-10=0
a = 3; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·3·(-10)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*3}=\frac{-12}{6} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*3}=\frac{10}{6} =1+2/3 $
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