2.25t^2-38t-38=0

Solution for 2.25t^2-38t-38=0 equation:

2.25t^2-38t-38=0

a = 2.25; b = -38; c = -38;
Δ = b2-4ac
Δ = -382-4·2.25·(-38)
Δ = 1786
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{-(-38)-\sqrt{1786}}{2*2.25}=\frac{38-\sqrt{1786}}{4.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+\sqrt{1786}}{2*2.25}=\frac{38+\sqrt{1786}}{4.5}
$