3t^2-+19t+16=0

Solution for 3t^2-+19t+16=0 equation:

3t^2-+19t+16=0

We add all the numbers together, and all the variables

3t^2+19t=0

a = 3; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·3·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*3}=\frac{-38}{6}
=-6+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*3}=\frac{0}{6}
=0
$