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3a^2+20a-68=0
a = 3; b = 20; c = -68;
Δ = b2-4ac
Δ = 202-4·3·(-68)
Δ = 1216
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{19}}{2*3}=\frac{-20-8\sqrt{19}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{19}}{2*3}=\frac{-20+8\sqrt{19}}{6} $
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