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3a^2-18a-98=0
a = 3; b = -18; c = -98;
Δ = b2-4ac
Δ = -182-4·3·(-98)
Δ = 1500
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1500}=\sqrt{100*15}=\sqrt{100}*\sqrt{15}=10\sqrt{15}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-10\sqrt{15}}{2*3}=\frac{18-10\sqrt{15}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+10\sqrt{15}}{2*3}=\frac{18+10\sqrt{15}}{6} $
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