16=4+3(t2)

Solution for 16=4+3(t2) equation:

16=4+3(t2)

We move all terms to the left:

16-(4+3(t2))=0

We add all the numbers together, and all the variables

-(+3t^2+4)+16=0

We get rid of parentheses

-3t^2-4+16=0

We add all the numbers together, and all the variables

-3t^2+12=0

a = -3; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-3)·12
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-3}=\frac{-12}{-6}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-3}=\frac{12}{-6}
=-2
$