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5x^2-216=0
a = 5; b = 0; c = -216;
Δ = b2-4ac
Δ = 02-4·5·(-216)
Δ = 4320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4320}=\sqrt{144*30}=\sqrt{144}*\sqrt{30}=12\sqrt{30}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{30}}{2*5}=\frac{0-12\sqrt{30}}{10} =-\frac{12\sqrt{30}}{10} =-\frac{6\sqrt{30}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{30}}{2*5}=\frac{0+12\sqrt{30}}{10} =\frac{12\sqrt{30}}{10} =\frac{6\sqrt{30}}{5} $
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