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x^2+3x-625=0
a = 1; b = 3; c = -625;
Δ = b2-4ac
Δ = 32-4·1·(-625)
Δ = 2509
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{-(3)-\sqrt{2509}}{2*1}=\frac{-3-\sqrt{2509}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{2509}}{2*1}=\frac{-3+\sqrt{2509}}{2} $
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