4t-2+t^2=6+t

Solution for 4t-2+t^2=6+t equation:

4t-2+t^2=6+t

We move all terms to the left:

4t-2+t^2-(6+t)=0

We add all the numbers together, and all the variables

t^2+4t-(t+6)-2=0

We get rid of parentheses

t^2+4t-t-6-2=0

We add all the numbers together, and all the variables

t^2+3t-8=0

a = 1; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·1·(-8)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{41}}{2*1}=\frac{-3-\sqrt{41}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*1}=\frac{-3+\sqrt{41}}{2}
$