40=5t+t^2

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

40=5t+t^2

We move all terms to the left:

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

We get rid of parentheses

-t^2-5t+40=0

We add all the numbers together, and all the variables

-1t^2-5t+40=0

a = -1; b = -5; c = +40;
Δ = b2-4ac
Δ = -52-4·(-1)·40
Δ = 185
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{185}}{2*-1}=\frac{5-\sqrt{185}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{185}}{2*-1}=\frac{5+\sqrt{185}}{-2}
$