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3x^2-40x-41=0
a = 3; b = -40; c = -41;
Δ = b2-4ac
Δ = -402-4·3·(-41)
Δ = 2092
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2092}=\sqrt{4*523}=\sqrt{4}*\sqrt{523}=2\sqrt{523}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{523}}{2*3}=\frac{40-2\sqrt{523}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{523}}{2*3}=\frac{40+2\sqrt{523}}{6} $
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