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25x^2+5x-1269=0
a = 25; b = 5; c = -1269;
Δ = b2-4ac
Δ = 52-4·25·(-1269)
Δ = 126925
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{126925}=\sqrt{25*5077}=\sqrt{25}*\sqrt{5077}=5\sqrt{5077}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{5077}}{2*25}=\frac{-5-5\sqrt{5077}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{5077}}{2*25}=\frac{-5+5\sqrt{5077}}{50} $
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