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31x^2+40x-31=0
a = 31; b = 40; c = -31;
Δ = b2-4ac
Δ = 402-4·31·(-31)
Δ = 5444
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5444}=\sqrt{4*1361}=\sqrt{4}*\sqrt{1361}=2\sqrt{1361}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{1361}}{2*31}=\frac{-40-2\sqrt{1361}}{62} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{1361}}{2*31}=\frac{-40+2\sqrt{1361}}{62} $
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