48=10t+2t^2

Solution for 48=10t+2t^2 equation:

48=10t+2t^2

We move all terms to the left:

48-(10t+2t^2)=0

We get rid of parentheses

-2t^2-10t+48=0

a = -2; b = -10; c = +48;
Δ = b2-4ac
Δ = -102-4·(-2)·48
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*-2}=\frac{-12}{-4}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*-2}=\frac{32}{-4}
=-8
$