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16x^2=3
We move all terms to the left:
16x^2-(3)=0
a = 16; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·16·(-3)
Δ = 192
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*16}=\frac{0-8\sqrt{3}}{32} =-\frac{8\sqrt{3}}{32} =-\frac{\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*16}=\frac{0+8\sqrt{3}}{32} =\frac{8\sqrt{3}}{32} =\frac{\sqrt{3}}{4} $
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