0t=4t2+48t+3

Solution for 0t=4t2+48t+3 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4t^2-47t-3=0

a = -4; b = -47; c = -3;
Δ = b2-4ac
Δ = -472-4·(-4)·(-3)
Δ = 2161
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{-(-47)-\sqrt{2161}}{2*-4}=\frac{47-\sqrt{2161}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{2161}}{2*-4}=\frac{47+\sqrt{2161}}{-8}
$