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12x^2-16=0
a = 12; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·12·(-16)
Δ = 768
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*12}=\frac{0-16\sqrt{3}}{24} =-\frac{16\sqrt{3}}{24} =-\frac{2\sqrt{3}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*12}=\frac{0+16\sqrt{3}}{24} =\frac{16\sqrt{3}}{24} =\frac{2\sqrt{3}}{3} $
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