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29x^2-16=0
a = 29; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·29·(-16)
Δ = 1856
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1856}=\sqrt{64*29}=\sqrt{64}*\sqrt{29}=8\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{29}}{2*29}=\frac{0-8\sqrt{29}}{58} =-\frac{8\sqrt{29}}{58} =-\frac{4\sqrt{29}}{29} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{29}}{2*29}=\frac{0+8\sqrt{29}}{58} =\frac{8\sqrt{29}}{58} =\frac{4\sqrt{29}}{29} $
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