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x^2+25x+27=0
a = 1; b = 25; c = +27;
Δ = b2-4ac
Δ = 252-4·1·27
Δ = 517
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{517}}{2*1}=\frac{-25-\sqrt{517}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{517}}{2*1}=\frac{-25+\sqrt{517}}{2} $
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