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8x^2-19x=2x^2-x
We move all terms to the left:
8x^2-19x-(2x^2-x)=0
We get rid of parentheses
8x^2-2x^2-19x+x=0
We add all the numbers together, and all the variables
6x^2-18x=0
a = 6; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·6·0
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*6}=\frac{0}{12} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*6}=\frac{36}{12} =3 $
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