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(t+4)(t+2)=3
We move all terms to the left:
(t+4)(t+2)-(3)=0
We multiply parentheses ..
(+t^2+2t+4t+8)-3=0
We get rid of parentheses
t^2+2t+4t+8-3=0
We add all the numbers together, and all the variables
t^2+6t+5=0
a = 1; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·1·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*1}=\frac{-10}{2} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*1}=\frac{-2}{2} =-1 $
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