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x^2+2x=306
We move all terms to the left:
x^2+2x-(306)=0
a = 1; b = 2; c = -306;
Δ = b2-4ac
Δ = 22-4·1·(-306)
Δ = 1228
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1228}=\sqrt{4*307}=\sqrt{4}*\sqrt{307}=2\sqrt{307}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{307}}{2*1}=\frac{-2-2\sqrt{307}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{307}}{2*1}=\frac{-2+2\sqrt{307}}{2} $
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