t^2+1.5t-45=0

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

t^2+1.5t-45=0

a = 1; b = 1.5; c = -45;
Δ = b2-4ac
Δ = 1.52-4·1·(-45)
Δ = 182.25
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{-(1.5)-\sqrt{182.25}}{2*1}=\frac{-1.5-\sqrt{182.25}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{182.25}}{2*1}=\frac{-1.5+\sqrt{182.25}}{2}
$