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-8=3+x/2x^2
We move all terms to the left:
-8-(3+x/2x^2)=0
Domain of the equation: 2x^2)!=0We get rid of parentheses
x!=0/1
x!=0
x∈R
-x/2x^2-3-8=0
We multiply all the terms by the denominator
-x-3*2x^2-8*2x^2=0
We add all the numbers together, and all the variables
-1x-3*2x^2-8*2x^2=0
Wy multiply elements
-6x^2-16x^2-1x=0
We add all the numbers together, and all the variables
-22x^2-1x=0
a = -22; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-22)·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-22}=\frac{0}{-44} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-22}=\frac{2}{-44} =-1/22 $
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