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3b^2-12b+2=0
a = 3; b = -12; c = +2;
Δ = b2-4ac
Δ = -122-4·3·2
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{30}}{2*3}=\frac{12-2\sqrt{30}}{6} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{30}}{2*3}=\frac{12+2\sqrt{30}}{6} $
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