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x^2+50x+225=0
a = 1; b = 50; c = +225;
Δ = b2-4ac
Δ = 502-4·1·225
Δ = 1600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-40}{2*1}=\frac{-90}{2} =-45 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+40}{2*1}=\frac{-10}{2} =-5 $
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