t^2+6t+8=t+2

Solution for t^2+6t+8=t+2 equation:

t^2+6t+8=t+2

We move all terms to the left:

t^2+6t+8-(t+2)=0

We get rid of parentheses

t^2+6t-t-2+8=0

We add all the numbers together, and all the variables

t^2+5t+6=0

a = 1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*1}=\frac{-6}{2}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*1}=\frac{-4}{2}
=-2
$