16t^2+72t-2.25=0

Solution for 16t^2+72t-2.25=0 equation:

16t^2+72t-2.25=0

a = 16; b = 72; c = -2.25;
Δ = b2-4ac
Δ = 722-4·16·(-2.25)
Δ = 5328
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:

$\sqrt{\Delta}=\sqrt{5328}=\sqrt{144*37}=\sqrt{144}*\sqrt{37}=12\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-12\sqrt{37}}{2*16}=\frac{-72-12\sqrt{37}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+12\sqrt{37}}{2*16}=\frac{-72+12\sqrt{37}}{32}
$