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5x^2-4=0
a = 5; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·5·(-4)
Δ = 80
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*5}=\frac{0-4\sqrt{5}}{10} =-\frac{4\sqrt{5}}{10} =-\frac{2\sqrt{5}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*5}=\frac{0+4\sqrt{5}}{10} =\frac{4\sqrt{5}}{10} =\frac{2\sqrt{5}}{5} $
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