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8=4x^2-4x
We move all terms to the left:
8-(4x^2-4x)=0
We get rid of parentheses
-4x^2+4x+8=0
a = -4; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-4)·8
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*-4}=\frac{-16}{-8} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*-4}=\frac{8}{-8} =-1 $
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