-1.5t^2+45t-4=0

Solution for -1.5t^2+45t-4=0 equation:

-1.5t^2+45t-4=0

a = -1.5; b = 45; c = -4;
Δ = b2-4ac
Δ = 452-4·(-1.5)·(-4)
Δ = 2001
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{-(45)-\sqrt{2001}}{2*-1.5}=\frac{-45-\sqrt{2001}}{-3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{2001}}{2*-1.5}=\frac{-45+\sqrt{2001}}{-3}
$