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10X^2-5X-491=0
a = 10; b = -5; c = -491;
Δ = b2-4ac
Δ = -52-4·10·(-491)
Δ = 19665
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{19665}=\sqrt{9*2185}=\sqrt{9}*\sqrt{2185}=3\sqrt{2185}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3\sqrt{2185}}{2*10}=\frac{5-3\sqrt{2185}}{20} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3\sqrt{2185}}{2*10}=\frac{5+3\sqrt{2185}}{20} $
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