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40x^2+6x-1=0
a = 40; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·40·(-1)
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*40}=\frac{-20}{80} =-1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*40}=\frac{8}{80} =1/10 $
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