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3u^2-10u+4=0
a = 3; b = -10; c = +4;
Δ = b2-4ac
Δ = -102-4·3·4
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{13}}{2*3}=\frac{10-2\sqrt{13}}{6} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{13}}{2*3}=\frac{10+2\sqrt{13}}{6} $
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