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