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X^2+2X=216
We move all terms to the left:
X^2+2X-(216)=0
a = 1; b = 2; c = -216;
Δ = b2-4ac
Δ = 22-4·1·(-216)
Δ = 868
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{217}}{2*1}=\frac{-2-2\sqrt{217}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{217}}{2*1}=\frac{-2+2\sqrt{217}}{2} $
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