-2t^2+5t+6=t

Solution for -2t^2+5t+6=t equation:

-2t^2+5t+6=t

We move all terms to the left:

-2t^2+5t+6-(t)=0

We add all the numbers together, and all the variables

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

a = -2; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-2)·6
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-2}=\frac{-12}{-4}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-2}=\frac{4}{-4}
=-1
$