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(x)=2x^2-4x
We move all terms to the left:
(x)-(2x^2-4x)=0
We get rid of parentheses
-2x^2+x+4x=0
We add all the numbers together, and all the variables
-2x^2+5x=0
a = -2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-2)·0
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-2}=\frac{-10}{-4} =2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-2}=\frac{0}{-4} =0 $
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