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x^2+2x+7=410
We move all terms to the left:
x^2+2x+7-(410)=0
We add all the numbers together, and all the variables
x^2+2x-403=0
a = 1; b = 2; c = -403;
Δ = b2-4ac
Δ = 22-4·1·(-403)
Δ = 1616
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1616}=\sqrt{16*101}=\sqrt{16}*\sqrt{101}=4\sqrt{101}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{101}}{2*1}=\frac{-2-4\sqrt{101}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{101}}{2*1}=\frac{-2+4\sqrt{101}}{2} $
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