.29t/t^2+16=0

Solution for .29t/t^2+16=0 equation:

.29t/t^2+16=0


Domain of the equation: t^2!=0

t^2!=0/

t^2!=√0

t!=0

t∈R


We multiply all the terms by the denominator


.29t+16*t^2=0

We add all the numbers together, and all the variables

16t^2+.29t=0

a = 16; b = .29; c = 0;
Δ = b2-4ac
Δ = .292-4·16·0
Δ = 0.0841
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{-(.29)-\sqrt{0.0841}}{2*16}=\frac{-0.29-\sqrt{0.0841}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.29)+\sqrt{0.0841}}{2*16}=\frac{-0.29+\sqrt{0.0841}}{32}
$