0=2t+t^2-48

Solution for 0=2t+t^2-48 equation:

0=2t+t^2-48

We move all terms to the left:

0-(2t+t^2-48)=0

We add all the numbers together, and all the variables

-(2t+t^2-48)=0

We get rid of parentheses

-t^2-2t+48=0

We add all the numbers together, and all the variables

-1t^2-2t+48=0

a = -1; b = -2; c = +48;
Δ = b2-4ac
Δ = -22-4·(-1)·48
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*-1}=\frac{-12}{-2}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*-1}=\frac{16}{-2}
=-8
$