t2+9t+14=(t+2)(t+)

Solution for t2+9t+14=(t+2)(t+) equation:

t2+9t+14=(t+2)(t+)

We move all terms to the left:

t2+9t+14-((t+2)(t+))=0

We add all the numbers together, and all the variables

t2+9t-((t+2)(+t))+14=0

We add all the numbers together, and all the variables

t^2+9t-((t+2)(+t))+14=0

We multiply parentheses ..

t^2-((+t^2+2t))+9t+14=0


We calculate terms in parentheses: -((+t^2+2t)), so:

(+t^2+2t)

We get rid of parentheses

t^2+2t

Back to the equation:

-(t^2+2t)


We add all the numbers together, and all the variables

t^2+9t-(t^2+2t)+14=0

We get rid of parentheses

t^2-t^2+9t-2t+14=0

We add all the numbers together, and all the variables

7t+14=0

We move all terms containing t to the left, all other terms to the right

7t=-14

t=-14/7

t=-2