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25x^2-54x+3=0
a = 25; b = -54; c = +3;
Δ = b2-4ac
Δ = -542-4·25·3
Δ = 2616
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2616}=\sqrt{4*654}=\sqrt{4}*\sqrt{654}=2\sqrt{654}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-54)-2\sqrt{654}}{2*25}=\frac{54-2\sqrt{654}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-54)+2\sqrt{654}}{2*25}=\frac{54+2\sqrt{654}}{50} $
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