40=-5t^2+30t

Solution for 40=-5t^2+30t equation:

40=-5t^2+30t

We move all terms to the left:

40-(-5t^2+30t)=0

We get rid of parentheses

5t^2-30t+40=0

a = 5; b = -30; c = +40;
Δ = b2-4ac
Δ = -302-4·5·40
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-10}{2*5}=\frac{20}{10}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+10}{2*5}=\frac{40}{10}
=4
$