2(t+1)=t^2

Solution for 2(t+1)=t^2 equation:

2(t+1)=t^2

We move all terms to the left:

2(t+1)-(t^2)=0

determiningTheFunctionDomain
-t^2+2(t+1)=0

We add all the numbers together, and all the variables

-1t^2+2(t+1)=0

We multiply parentheses

-1t^2+2t+2=0

a = -1; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-1)·2
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*-1}=\frac{-2-2\sqrt{3}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*-1}=\frac{-2+2\sqrt{3}}{-2}
$