t=16t^2+148t

Solution for t=16t^2+148t equation:

t=16t^2+148t

We move all terms to the left:

t-(16t^2+148t)=0

We get rid of parentheses

-16t^2+t-148t=0

We add all the numbers together, and all the variables

-16t^2-147t=0

a = -16; b = -147; c = 0;
Δ = b2-4ac
Δ = -1472-4·(-16)·0
Δ = 21609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{21609}=147$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-147)-147}{2*-16}=\frac{0}{-32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-147)+147}{2*-16}=\frac{294}{-32}
=-9+3/16
$