48=17t+3t^2

Solution for 48=17t+3t^2 equation:

48=17t+3t^2

We move all terms to the left:

48-(17t+3t^2)=0

We get rid of parentheses

-3t^2-17t+48=0

a = -3; b = -17; c = +48;
Δ = b2-4ac
Δ = -172-4·(-3)·48
Δ = 865
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{-(-17)-\sqrt{865}}{2*-3}=\frac{17-\sqrt{865}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{865}}{2*-3}=\frac{17+\sqrt{865}}{-6}
$