t2-4t=3t2

Solution for t2-4t=3t2 equation:

t2-4t=3t^2

We move all terms to the left:

t2-4t-(3t^2)=0

determiningTheFunctionDomain
-3t^2+t2-4t=0

We add all the numbers together, and all the variables

-2t^2-4t=0

a = -2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-2)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-2}=\frac{0}{-4}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-2}=\frac{8}{-4}
=-2
$