12(t)=16t^2+32t+5

Solution for 12(t)=16t^2+32t+5 equation:

12(t)=16t^2+32t+5

We move all terms to the left:

12(t)-(16t^2+32t+5)=0

We get rid of parentheses

-16t^2+12t-32t-5=0

We add all the numbers together, and all the variables

-16t^2-20t-5=0

a = -16; b = -20; c = -5;
Δ = b2-4ac
Δ = -202-4·(-16)·(-5)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{5}}{2*-16}=\frac{20-4\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{5}}{2*-16}=\frac{20+4\sqrt{5}}{-32}
$