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3b^2=316+22
We move all terms to the left:
3b^2-(316+22)=0
We add all the numbers together, and all the variables
3b^2-338=0
a = 3; b = 0; c = -338;
Δ = b2-4ac
Δ = 02-4·3·(-338)
Δ = 4056
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4056}=\sqrt{676*6}=\sqrt{676}*\sqrt{6}=26\sqrt{6}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-26\sqrt{6}}{2*3}=\frac{0-26\sqrt{6}}{6} =-\frac{26\sqrt{6}}{6} =-\frac{13\sqrt{6}}{3} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+26\sqrt{6}}{2*3}=\frac{0+26\sqrt{6}}{6} =\frac{26\sqrt{6}}{6} =\frac{13\sqrt{6}}{3} $
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