t2+2t-0,16=0

Solution for t2+2t-0,16=0 equation:

t2+2t-0.16=0

We add all the numbers together, and all the variables

t^2+2t-0.16=0

a = 1; b = 2; c = -0.16;
Δ = b2-4ac
Δ = 22-4·1·(-0.16)
Δ = 4.64
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)-\sqrt{4.64}}{2*1}=\frac{-2-\sqrt{4.64}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{4.64}}{2*1}=\frac{-2+\sqrt{4.64}}{2}
$