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40x^2-53x+6=0
a = 40; b = -53; c = +6;
Δ = b2-4ac
Δ = -532-4·40·6
Δ = 1849
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{1849}=43$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-43}{2*40}=\frac{10}{80} =1/8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+43}{2*40}=\frac{96}{80} =1+1/5 $
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