3(t2-9)2+16(t2-9)=-5

Solution for 3(t2-9)2+16(t2-9)=-5 equation:

3(t2-9)2+16(t2-9)=-5

We move all terms to the left:

3(t2-9)2+16(t2-9)-(-5)=0

We add all the numbers together, and all the variables

3(+t^2-9)2+16(+t^2-9)-(-5)=0

We add all the numbers together, and all the variables

3(+t^2-9)2+16(+t^2-9)+5=0

We multiply parentheses

6t^2+16t^2-54-144+5=0

We add all the numbers together, and all the variables

22t^2-193=0

a = 22; b = 0; c = -193;
Δ = b2-4ac
Δ = 02-4·22·(-193)
Δ = 16984
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{16984}=\sqrt{4*4246}=\sqrt{4}*\sqrt{4246}=2\sqrt{4246}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{4246}}{2*22}=\frac{0-2\sqrt{4246}}{44}
=-\frac{2\sqrt{4246}}{44}
=-\frac{\sqrt{4246}}{22}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{4246}}{2*22}=\frac{0+2\sqrt{4246}}{44}
=\frac{2\sqrt{4246}}{44}
=\frac{\sqrt{4246}}{22}
$