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5x^2+20x-6000=0
a = 5; b = 20; c = -6000;
Δ = b2-4ac
Δ = 202-4·5·(-6000)
Δ = 120400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120400}=\sqrt{400*301}=\sqrt{400}*\sqrt{301}=20\sqrt{301}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{301}}{2*5}=\frac{-20-20\sqrt{301}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{301}}{2*5}=\frac{-20+20\sqrt{301}}{10} $
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