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2x^2+2x-508=0
a = 2; b = 2; c = -508;
Δ = b2-4ac
Δ = 22-4·2·(-508)
Δ = 4068
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4068}=\sqrt{36*113}=\sqrt{36}*\sqrt{113}=6\sqrt{113}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{113}}{2*2}=\frac{-2-6\sqrt{113}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{113}}{2*2}=\frac{-2+6\sqrt{113}}{4} $
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