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12x^2-50=0
a = 12; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·12·(-50)
Δ = 2400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{6}}{2*12}=\frac{0-20\sqrt{6}}{24} =-\frac{20\sqrt{6}}{24} =-\frac{5\sqrt{6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{6}}{2*12}=\frac{0+20\sqrt{6}}{24} =\frac{20\sqrt{6}}{24} =\frac{5\sqrt{6}}{6} $
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