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12b^2-18=0
a = 12; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·12·(-18)
Δ = 864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{6}}{2*12}=\frac{0-12\sqrt{6}}{24} =-\frac{12\sqrt{6}}{24} =-\frac{\sqrt{6}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{6}}{2*12}=\frac{0+12\sqrt{6}}{24} =\frac{12\sqrt{6}}{24} =\frac{\sqrt{6}}{2} $
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