10+40t-5t^2=0

Solution for 10+40t-5t^2=0 equation:

10+40t-5t^2=0

a = -5; b = 40; c = +10;
Δ = b2-4ac
Δ = 402-4·(-5)·10
Δ = 1800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-30\sqrt{2}}{2*-5}=\frac{-40-30\sqrt{2}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+30\sqrt{2}}{2*-5}=\frac{-40+30\sqrt{2}}{-10}
$