t^2-4.49t+2.27=0

Solution for t^2-4.49t+2.27=0 equation:

t^2-4.49t+2.27=0

a = 1; b = -4.49; c = +2.27;
Δ = b2-4ac
Δ = -4.492-4·1·2.27
Δ = 11.0801
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{-(-4.49)-\sqrt{11.0801}}{2*1}=\frac{4.49-\sqrt{11.0801}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4.49)+\sqrt{11.0801}}{2*1}=\frac{4.49+\sqrt{11.0801}}{2}
$