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2x^2+2x+1=208
We move all terms to the left:
2x^2+2x+1-(208)=0
We add all the numbers together, and all the variables
2x^2+2x-207=0
a = 2; b = 2; c = -207;
Δ = b2-4ac
Δ = 22-4·2·(-207)
Δ = 1660
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1660}=\sqrt{4*415}=\sqrt{4}*\sqrt{415}=2\sqrt{415}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{415}}{2*2}=\frac{-2-2\sqrt{415}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{415}}{2*2}=\frac{-2+2\sqrt{415}}{4} $
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