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40x^2-56x+16=0
a = 40; b = -56; c = +16;
Δ = b2-4ac
Δ = -562-4·40·16
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-24}{2*40}=\frac{32}{80} =2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+24}{2*40}=\frac{80}{80} =1 $
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