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3b^2+2b-21=0
a = 3; b = 2; c = -21;
Δ = b2-4ac
Δ = 22-4·3·(-21)
Δ = 256
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{256}=16$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-16}{2*3}=\frac{-18}{6} =-3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+16}{2*3}=\frac{14}{6} =2+1/3 $
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