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a^2+6a-216=0
a = 1; b = 6; c = -216;
Δ = b2-4ac
Δ = 62-4·1·(-216)
Δ = 900
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{900}=30$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-30}{2*1}=\frac{-36}{2} =-18 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+30}{2*1}=\frac{24}{2} =12 $
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