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40x^2-60x+20=0
a = 40; b = -60; c = +20;
Δ = b2-4ac
Δ = -602-4·40·20
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-20}{2*40}=\frac{40}{80} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+20}{2*40}=\frac{80}{80} =1 $
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