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20x^2+40x-60=0
a = 20; b = 40; c = -60;
Δ = b2-4ac
Δ = 402-4·20·(-60)
Δ = 6400
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{6400}=80$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-80}{2*20}=\frac{-120}{40} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+80}{2*20}=\frac{40}{40} =1 $
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