(-18(t^2-100))/(t^2+5*t+100)^2=0

Solution for (-18(t^2-100))/(t^2+5*t+100)^2=0 equation:

(-18(t^2-100))/(t^2+5t+100)^2=0


Domain of the equation: (t^2+5t+100)^2!=0

t∈R


We multiply all the terms by the denominator


(-18(t^2-100))=0


We calculate terms in parentheses: +(-18(t^2-100)), so:

-18(t^2-100)

We multiply parentheses

-18t^2+1800

Back to the equation:

+(-18t^2+1800)


We get rid of parentheses

-18t^2+1800=0

a = -18; b = 0; c = +1800;
Δ = b2-4ac
Δ = 02-4·(-18)·1800
Δ = 129600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{129600}=360$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-360}{2*-18}=\frac{-360}{-36}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+360}{2*-18}=\frac{360}{-36}
=-10
$