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3b^2=192
We move all terms to the left:
3b^2-(192)=0
a = 3; b = 0; c = -192;
Δ = b2-4ac
Δ = 02-4·3·(-192)
Δ = 2304
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*3}=\frac{-48}{6} =-8 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*3}=\frac{48}{6} =8 $
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