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