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2x^2-192=0
a = 2; b = 0; c = -192;
Δ = b2-4ac
Δ = 02-4·2·(-192)
Δ = 1536
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{6}}{2*2}=\frac{0-16\sqrt{6}}{4} =-\frac{16\sqrt{6}}{4} =-4\sqrt{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{6}}{2*2}=\frac{0+16\sqrt{6}}{4} =\frac{16\sqrt{6}}{4} =4\sqrt{6} $
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