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b(b+6)=112
We move all terms to the left:
b(b+6)-(112)=0
We multiply parentheses
b^2+6b-112=0
a = 1; b = 6; c = -112;
Δ = b2-4ac
Δ = 62-4·1·(-112)
Δ = 484
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{484}=22$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-22}{2*1}=\frac{-28}{2} =-14 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+22}{2*1}=\frac{16}{2} =8 $
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