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4=2x^2+2x
We move all terms to the left:
4-(2x^2+2x)=0
We get rid of parentheses
-2x^2-2x+4=0
a = -2; b = -2; c = +4;
Δ = b2-4ac
Δ = -22-4·(-2)·4
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*-2}=\frac{-4}{-4} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-2}=\frac{8}{-4} =-2 $
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