t2-40t+300=0

Solution for t2-40t+300=0 equation:

t2-40t+300=0

We add all the numbers together, and all the variables

t^2-40t+300=0

a = 1; b = -40; c = +300;
Δ = b2-4ac
Δ = -402-4·1·300
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20}{2*1}=\frac{20}{2}
=10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20}{2*1}=\frac{60}{2}
=30
$