t^2+48=14t

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

t^2+48=14t

We move all terms to the left:

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

a = 1; b = -14; c = +48;
Δ = b2-4ac
Δ = -142-4·1·48
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2}{2*1}=\frac{12}{2}
=6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*1}=\frac{16}{2}
=8
$