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3a^2-20a+8=0
a = 3; b = -20; c = +8;
Δ = b2-4ac
Δ = -202-4·3·8
Δ = 304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{19}}{2*3}=\frac{20-4\sqrt{19}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{19}}{2*3}=\frac{20+4\sqrt{19}}{6} $
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