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q(q+11)+12(q+11)=0
We multiply parentheses
q^2+11q+12q+132=0
We add all the numbers together, and all the variables
q^2+23q+132=0
a = 1; b = 23; c = +132;
Δ = b2-4ac
Δ = 232-4·1·132
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-1}{2*1}=\frac{-24}{2} =-12 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+1}{2*1}=\frac{-22}{2} =-11 $
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