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12x^2=4x+6x^2
We move all terms to the left:
12x^2-(4x+6x^2)=0
We get rid of parentheses
12x^2-6x^2-4x=0
We add all the numbers together, and all the variables
6x^2-4x=0
a = 6; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·6·0
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*6}=\frac{0}{12} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*6}=\frac{8}{12} =2/3 $
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