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3b^2-147=0
a = 3; b = 0; c = -147;
Δ = b2-4ac
Δ = 02-4·3·(-147)
Δ = 1764
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{1764}=42$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*3}=\frac{-42}{6} =-7 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*3}=\frac{42}{6} =7 $
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