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5n^2-25=0
a = 5; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·5·(-25)
Δ = 500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{5}}{2*5}=\frac{0-10\sqrt{5}}{10} =-\frac{10\sqrt{5}}{10} =-\sqrt{5} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{5}}{2*5}=\frac{0+10\sqrt{5}}{10} =\frac{10\sqrt{5}}{10} =\sqrt{5} $
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