3(t2-16)2+19(t2-16)=-6

Solution for 3(t2-16)2+19(t2-16)=-6 equation:

3(t2-16)2+19(t2-16)=-6

We move all terms to the left:

3(t2-16)2+19(t2-16)-(-6)=0

We add all the numbers together, and all the variables

3(+t^2-16)2+19(+t^2-16)-(-6)=0

We add all the numbers together, and all the variables

3(+t^2-16)2+19(+t^2-16)+6=0

We multiply parentheses

6t^2+19t^2-96-304+6=0

We add all the numbers together, and all the variables

25t^2-394=0

a = 25; b = 0; c = -394;
Δ = b2-4ac
Δ = 02-4·25·(-394)
Δ = 39400
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:

$\sqrt{\Delta}=\sqrt{39400}=\sqrt{100*394}=\sqrt{100}*\sqrt{394}=10\sqrt{394}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{394}}{2*25}=\frac{0-10\sqrt{394}}{50}
=-\frac{10\sqrt{394}}{50}
=-\frac{\sqrt{394}}{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{394}}{2*25}=\frac{0+10\sqrt{394}}{50}
=\frac{10\sqrt{394}}{50}
=\frac{\sqrt{394}}{5}
$