3=t^2-14t+48

Solution for 3=t^2-14t+48 equation:

3=t^2-14t+48

We move all terms to the left:

3-(t^2-14t+48)=0

We get rid of parentheses

-t^2+14t-48+3=0

We add all the numbers together, and all the variables

-1t^2+14t-45=0

a = -1; b = 14; c = -45;
Δ = b2-4ac
Δ = 142-4·(-1)·(-45)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*-1}=\frac{-18}{-2}
=+9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*-1}=\frac{-10}{-2}
=+5
$