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5x^2+40x-500=0
a = 5; b = 40; c = -500;
Δ = b2-4ac
Δ = 402-4·5·(-500)
Δ = 11600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11600}=\sqrt{400*29}=\sqrt{400}*\sqrt{29}=20\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{29}}{2*5}=\frac{-40-20\sqrt{29}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{29}}{2*5}=\frac{-40+20\sqrt{29}}{10} $
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