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5x^2+50x-2000=0
a = 5; b = 50; c = -2000;
Δ = b2-4ac
Δ = 502-4·5·(-2000)
Δ = 42500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{42500}=\sqrt{2500*17}=\sqrt{2500}*\sqrt{17}=50\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50\sqrt{17}}{2*5}=\frac{-50-50\sqrt{17}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50\sqrt{17}}{2*5}=\frac{-50+50\sqrt{17}}{10} $
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