16-t^2=4/3

Solution for 16-t^2=4/3 equation:

16-t^2=4/3

We move all terms to the left:

16-t^2-(4/3)=0

We add all the numbers together, and all the variables

-t^2+16-(+4/3)=0

We add all the numbers together, and all the variables

-1t^2+16-(+4/3)=0

We get rid of parentheses

-1t^2+16-4/3=0

We multiply all the terms by the denominator


-1t^2*3-4+16*3=0

We add all the numbers together, and all the variables

-1t^2*3+44=0

Wy multiply elements

-3t^2+44=0

a = -3; b = 0; c = +44;
Δ = b2-4ac
Δ = 02-4·(-3)·44
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{33}}{2*-3}=\frac{0-4\sqrt{33}}{-6}
=-\frac{4\sqrt{33}}{-6}
=-\frac{2\sqrt{33}}{-3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{33}}{2*-3}=\frac{0+4\sqrt{33}}{-6}
=\frac{4\sqrt{33}}{-6}
=\frac{2\sqrt{33}}{-3}
$