40=-4.9t^2+28t+10

Solution for 40=-4.9t^2+28t+10 equation:

40=-4.9t^2+28t+10

We move all terms to the left:

40-(-4.9t^2+28t+10)=0

We get rid of parentheses

4.9t^2-28t-10+40=0

We add all the numbers together, and all the variables

4.9t^2-28t+30=0

a = 4.9; b = -28; c = +30;
Δ = b2-4ac
Δ = -282-4·4.9·30
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-14}{2*4.9}=\frac{14}{9.8}
=1+3.5/8.1666666666667
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+14}{2*4.9}=\frac{42}{9.8}
=4+2.8/9.8
$