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6x^2+4x=5x-3x^2
We move all terms to the left:
6x^2+4x-(5x-3x^2)=0
We get rid of parentheses
6x^2+3x^2-5x+4x=0
We add all the numbers together, and all the variables
9x^2-1x=0
a = 9; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·9·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*9}=\frac{0}{18} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*9}=\frac{2}{18} =1/9 $
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