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4(a)^2=2-2a
We move all terms to the left:
4(a)^2-(2-2a)=0
We add all the numbers together, and all the variables
4a^2-(-2a+2)=0
We get rid of parentheses
4a^2+2a-2=0
a = 4; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·4·(-2)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*4}=\frac{-8}{8} =-1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*4}=\frac{4}{8} =1/2 $
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