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12x^2-5=0
a = 12; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·12·(-5)
Δ = 240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*12}=\frac{0-4\sqrt{15}}{24} =-\frac{4\sqrt{15}}{24} =-\frac{\sqrt{15}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*12}=\frac{0+4\sqrt{15}}{24} =\frac{4\sqrt{15}}{24} =\frac{\sqrt{15}}{6} $
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