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X^2+2X=68
We move all terms to the left:
X^2+2X-(68)=0
a = 1; b = 2; c = -68;
Δ = b2-4ac
Δ = 22-4·1·(-68)
Δ = 276
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{69}}{2*1}=\frac{-2-2\sqrt{69}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{69}}{2*1}=\frac{-2+2\sqrt{69}}{2} $
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