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12x=2x^2+16
We move all terms to the left:
12x-(2x^2+16)=0
We get rid of parentheses
-2x^2+12x-16=0
a = -2; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·(-2)·(-16)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*-2}=\frac{-16}{-4} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*-2}=\frac{-8}{-4} =+2 $
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