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40x^2-156x-100=0
a = 40; b = -156; c = -100;
Δ = b2-4ac
Δ = -1562-4·40·(-100)
Δ = 40336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40336}=\sqrt{16*2521}=\sqrt{16}*\sqrt{2521}=4\sqrt{2521}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-156)-4\sqrt{2521}}{2*40}=\frac{156-4\sqrt{2521}}{80} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-156)+4\sqrt{2521}}{2*40}=\frac{156+4\sqrt{2521}}{80} $
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