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8x=2x^2+2
We move all terms to the left:
8x-(2x^2+2)=0
We get rid of parentheses
-2x^2+8x-2=0
a = -2; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·(-2)·(-2)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{3}}{2*-2}=\frac{-8-4\sqrt{3}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{3}}{2*-2}=\frac{-8+4\sqrt{3}}{-4} $
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