4t^2+20t+2.5=0

Solution for 4t^2+20t+2.5=0 equation:

4t^2+20t+2.5=0

a = 4; b = 20; c = +2.5;
Δ = b2-4ac
Δ = 202-4·4·2.5
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6\sqrt{10}}{2*4}=\frac{-20-6\sqrt{10}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6\sqrt{10}}{2*4}=\frac{-20+6\sqrt{10}}{8}
$