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a^2+a=9
We move all terms to the left:
a^2+a-(9)=0
a = 1; b = 1; c = -9;
Δ = b2-4ac
Δ = 12-4·1·(-9)
Δ = 37
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{37}}{2*1}=\frac{-1-\sqrt{37}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{37}}{2*1}=\frac{-1+\sqrt{37}}{2} $
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