-t^2=22-(-16t^2+38)

Solution for -t^2=22-(-16t^2+38) equation:

-t^2=22-(-16t^2+38)

We move all terms to the left:

-t^2-(22-(-16t^2+38))=0

We add all the numbers together, and all the variables

-1t^2-(22-(-16t^2+38))=0


We calculate terms in parentheses: -(22-(-16t^2+38)), so:

22-(-16t^2+38)

determiningTheFunctionDomain
-(-16t^2+38)+22

We get rid of parentheses

16t^2-38+22

We add all the numbers together, and all the variables

16t^2-16

Back to the equation:

-(16t^2-16)


We get rid of parentheses

-1t^2-16t^2+16=0

We add all the numbers together, and all the variables

-17t^2+16=0

a = -17; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-17)·16
Δ = 1088
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:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{17}}{2*-17}=\frac{0-8\sqrt{17}}{-34}
=-\frac{8\sqrt{17}}{-34}
=-\frac{4\sqrt{17}}{-17}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{17}}{2*-17}=\frac{0+8\sqrt{17}}{-34}
=\frac{8\sqrt{17}}{-34}
=\frac{4\sqrt{17}}{-17}
$