t^2=2t^2+4t-96

Solution for t^2=2t^2+4t-96 equation:

t^2=2t^2+4t-96

We move all terms to the left:

t^2-(2t^2+4t-96)=0

We get rid of parentheses

t^2-2t^2-4t+96=0

We add all the numbers together, and all the variables

-1t^2-4t+96=0

a = -1; b = -4; c = +96;
Δ = b2-4ac
Δ = -42-4·(-1)·96
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*-1}=\frac{-16}{-2}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*-1}=\frac{24}{-2}
=-12
$