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36a^2-48=0
a = 36; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·36·(-48)
Δ = 6912
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6912}=\sqrt{2304*3}=\sqrt{2304}*\sqrt{3}=48\sqrt{3}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{3}}{2*36}=\frac{0-48\sqrt{3}}{72} =-\frac{48\sqrt{3}}{72} =-\frac{2\sqrt{3}}{3} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{3}}{2*36}=\frac{0+48\sqrt{3}}{72} =\frac{48\sqrt{3}}{72} =\frac{2\sqrt{3}}{3} $
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