48=t^2-4t+3

Solution for 48=t^2-4t+3 equation:

48=t^2-4t+3

We move all terms to the left:

48-(t^2-4t+3)=0

We get rid of parentheses

-t^2+4t-3+48=0

We add all the numbers together, and all the variables

-1t^2+4t+45=0

a = -1; b = 4; c = +45;
Δ = b2-4ac
Δ = 42-4·(-1)·45
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{-(4)-14}{2*-1}=\frac{-18}{-2}
=+9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*-1}=\frac{10}{-2}
=-5
$