0=-7t^2+28t+35

Solution for 0=-7t^2+28t+35 equation:

0=-7t^2+28t+35

We move all terms to the left:

0-(-7t^2+28t+35)=0

We add all the numbers together, and all the variables

-(-7t^2+28t+35)=0

We get rid of parentheses

7t^2-28t-35=0

a = 7; b = -28; c = -35;
Δ = b2-4ac
Δ = -282-4·7·(-35)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-42}{2*7}=\frac{-14}{14}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+42}{2*7}=\frac{70}{14}
=5
$