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x^2+50x-225=0
a = 1; b = 50; c = -225;
Δ = b2-4ac
Δ = 502-4·1·(-225)
Δ = 3400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3400}=\sqrt{100*34}=\sqrt{100}*\sqrt{34}=10\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{34}}{2*1}=\frac{-50-10\sqrt{34}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{34}}{2*1}=\frac{-50+10\sqrt{34}}{2} $
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