2t^2-22.5t=0

Solution for 2t^2-22.5t=0 equation:

2t^2-22.5t=0

a = 2; b = -22.5; c = 0;
Δ = b2-4ac
Δ = -22.52-4·2·0
Δ = 506.25
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{-(-22.5)-\sqrt{506.25}}{2*2}=\frac{22.5-\sqrt{506.25}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22.5)+\sqrt{506.25}}{2*2}=\frac{22.5+\sqrt{506.25}}{4}
$