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b^2+7b=30
We move all terms to the left:
b^2+7b-(30)=0
a = 1; b = 7; c = -30;
Δ = b2-4ac
Δ = 72-4·1·(-30)
Δ = 169
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{169}=13$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*1}=\frac{-20}{2} =-10 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*1}=\frac{6}{2} =3 $
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