7=16t^2+16t+3

Solution for 7=16t^2+16t+3 equation:

7=16t^2+16t+3

We move all terms to the left:

7-(16t^2+16t+3)=0

We get rid of parentheses

-16t^2-16t-3+7=0

We add all the numbers together, and all the variables

-16t^2-16t+4=0

a = -16; b = -16; c = +4;
Δ = b2-4ac
Δ = -162-4·(-16)·4
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{2}}{2*-16}=\frac{16-16\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{2}}{2*-16}=\frac{16+16\sqrt{2}}{-32}
$