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X^2+X=280
We move all terms to the left:
X^2+X-(280)=0
a = 1; b = 1; c = -280;
Δ = b2-4ac
Δ = 12-4·1·(-280)
Δ = 1121
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1121}}{2*1}=\frac{-1-\sqrt{1121}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1121}}{2*1}=\frac{-1+\sqrt{1121}}{2} $
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