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2x^2-2x=399
We move all terms to the left:
2x^2-2x-(399)=0
a = 2; b = -2; c = -399;
Δ = b2-4ac
Δ = -22-4·2·(-399)
Δ = 3196
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3196}=\sqrt{4*799}=\sqrt{4}*\sqrt{799}=2\sqrt{799}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{799}}{2*2}=\frac{2-2\sqrt{799}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{799}}{2*2}=\frac{2+2\sqrt{799}}{4} $
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