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245x^2-20x-8=0
a = 245; b = -20; c = -8;
Δ = b2-4ac
Δ = -202-4·245·(-8)
Δ = 8240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8240}=\sqrt{16*515}=\sqrt{16}*\sqrt{515}=4\sqrt{515}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{515}}{2*245}=\frac{20-4\sqrt{515}}{490} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{515}}{2*245}=\frac{20+4\sqrt{515}}{490} $
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