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25x^2-55x+25=0
a = 25; b = -55; c = +25;
Δ = b2-4ac
Δ = -552-4·25·25
Δ = 525
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{525}=\sqrt{25*21}=\sqrt{25}*\sqrt{21}=5\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-55)-5\sqrt{21}}{2*25}=\frac{55-5\sqrt{21}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+5\sqrt{21}}{2*25}=\frac{55+5\sqrt{21}}{50} $
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