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18a+a^2=72
We move all terms to the left:
18a+a^2-(72)=0
a = 1; b = 18; c = -72;
Δ = b2-4ac
Δ = 182-4·1·(-72)
Δ = 612
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{17}}{2*1}=\frac{-18-6\sqrt{17}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{17}}{2*1}=\frac{-18+6\sqrt{17}}{2} $
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