-9.4t2+5t+4=0

Solution for -9.4t2+5t+4=0 equation:

-9.4t^2+5t+4=0

a = -9.4; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-9.4)·4
Δ = 175.4
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{-(5)-\sqrt{175.4}}{2*-9.4}=\frac{-5-\sqrt{175.4}}{-18.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{175.4}}{2*-9.4}=\frac{-5+\sqrt{175.4}}{-18.8}
$