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0.4x^2-48x+1280=0
a = 0.4; b = -48; c = +1280;
Δ = b2-4ac
Δ = -482-4·0.4·1280
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16}{2*0.4}=\frac{32}{0.8} =40 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16}{2*0.4}=\frac{64}{0.8} =80 $
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