4,5t^2+20t-40=0

Solution for 4,5t^2+20t-40=0 equation:

4.5t^2+20t-40=0

a = 4.5; b = 20; c = -40;
Δ = b2-4ac
Δ = 202-4·4.5·(-40)
Δ = 1120
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:

$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{70}}{2*4.5}=\frac{-20-4\sqrt{70}}{9}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{70}}{2*4.5}=\frac{-20+4\sqrt{70}}{9}
$