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25x^2+25x-625=0
a = 25; b = 25; c = -625;
Δ = b2-4ac
Δ = 252-4·25·(-625)
Δ = 63125
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{63125}=\sqrt{625*101}=\sqrt{625}*\sqrt{101}=25\sqrt{101}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25\sqrt{101}}{2*25}=\frac{-25-25\sqrt{101}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25\sqrt{101}}{2*25}=\frac{-25+25\sqrt{101}}{50} $
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