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x^2-25x+3=0
a = 1; b = -25; c = +3;
Δ = b2-4ac
Δ = -252-4·1·3
Δ = 613
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{613}}{2*1}=\frac{25-\sqrt{613}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{613}}{2*1}=\frac{25+\sqrt{613}}{2} $
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