2.25t^2-18t-125=0

Solution for 2.25t^2-18t-125=0 equation:

2.25t^2-18t-125=0

a = 2.25; b = -18; c = -125;
Δ = b2-4ac
Δ = -182-4·2.25·(-125)
Δ = 1449
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:

$\sqrt{\Delta}=\sqrt{1449}=\sqrt{9*161}=\sqrt{9}*\sqrt{161}=3\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-3\sqrt{161}}{2*2.25}=\frac{18-3\sqrt{161}}{4.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+3\sqrt{161}}{2*2.25}=\frac{18+3\sqrt{161}}{4.5}
$