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4x^2+80x-500=0
a = 4; b = 80; c = -500;
Δ = b2-4ac
Δ = 802-4·4·(-500)
Δ = 14400
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{14400}=120$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-120}{2*4}=\frac{-200}{8} =-25 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+120}{2*4}=\frac{40}{8} =5 $
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