0=-16t^2+46t-28

Solution for 0=-16t^2+46t-28 equation:

0=-16t^2+46t-28

We move all terms to the left:

0-(-16t^2+46t-28)=0

We add all the numbers together, and all the variables

-(-16t^2+46t-28)=0

We get rid of parentheses

16t^2-46t+28=0

a = 16; b = -46; c = +28;
Δ = b2-4ac
Δ = -462-4·16·28
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-18}{2*16}=\frac{28}{32}
=7/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+18}{2*16}=\frac{64}{32}
=2
$