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x^2+12x=216
We move all terms to the left:
x^2+12x-(216)=0
a = 1; b = 12; c = -216;
Δ = b2-4ac
Δ = 122-4·1·(-216)
Δ = 1008
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{7}}{2*1}=\frac{-12-12\sqrt{7}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{7}}{2*1}=\frac{-12+12\sqrt{7}}{2} $
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