3t2+-16t+28=0

Solution for 3t2+-16t+28=0 equation:

3t^2+-16t+28=0

We add all the numbers together, and all the variables

3t^2-16t=0

a = 3; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·3·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*3}=\frac{0}{6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*3}=\frac{32}{6}
=5+1/3
$