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5x^2+25x=1
We move all terms to the left:
5x^2+25x-(1)=0
a = 5; b = 25; c = -1;
Δ = b2-4ac
Δ = 252-4·5·(-1)
Δ = 645
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{645}}{2*5}=\frac{-25-\sqrt{645}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{645}}{2*5}=\frac{-25+\sqrt{645}}{10} $
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