3=4t^2+8t+3

Solution for 3=4t^2+8t+3 equation:

3=4t^2+8t+3

We move all terms to the left:

3-(4t^2+8t+3)=0

We get rid of parentheses

-4t^2-8t-3+3=0

We add all the numbers together, and all the variables

-4t^2-8t=0

a = -4; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-4)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-4}=\frac{0}{-8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-4}=\frac{16}{-8}
=-2
$