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-10(x+3)=20(-1/2x+1)-2+12
We move all terms to the left:
-10(x+3)-(20(-1/2x+1)-2+12)=0
Domain of the equation: 2x+1)-2+12)!=0We multiply parentheses
x∈R
-10x-(20(-1/2x+1)-2+12)-30=0
We multiply all the terms by the denominator
-10x*2x-30*2x-1+1)-2+12)-(20(+1)-2+12)=0
We add all the numbers together, and all the variables
-10x*2x-30*2x-1+1)-2+12)-(201-2+12)=0
We add all the numbers together, and all the variables
-10x*2x-30*2x=0
Wy multiply elements
-20x^2-60x=0
a = -20; b = -60; c = 0;
Δ = b2-4ac
Δ = -602-4·(-20)·0
Δ = 3600
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{3600}=60x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-60}{2*-20}=\frac{0}{-40} =0x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+60}{2*-20}=\frac{120}{-40} =-3
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