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(s+9)(s+2)=-6
We move all terms to the left:
(s+9)(s+2)-(-6)=0
We add all the numbers together, and all the variables
(s+9)(s+2)+6=0
We multiply parentheses ..
(+s^2+2s+9s+18)+6=0
We get rid of parentheses
s^2+2s+9s+18+6=0
We add all the numbers together, and all the variables
s^2+11s+24=0
a = 1; b = 11; c = +24;
Δ = b2-4ac
Δ = 112-4·1·24
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*1}=\frac{-16}{2} =-8 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*1}=\frac{-6}{2} =-3 $
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