0.1t^2-2.6t+16=0

Solution for 0.1t^2-2.6t+16=0 equation:

0.1t^2-2.6t+16=0

a = 0.1; b = -2.6; c = +16;
Δ = b2-4ac
Δ = -2.62-4·0.1·16
Δ = 0.36
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{-(-2.6)-\sqrt{0.36}}{2*0.1}=\frac{2.6-\sqrt{0.36}}{0.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.6)+\sqrt{0.36}}{2*0.1}=\frac{2.6+\sqrt{0.36}}{0.2}
$