-(3/4)t=-12

Solution for -(3/4)t=-12 equation:

-(3/4)t=-12

We move all terms to the left:

-(3/4)t-(-12)=0


Domain of the equation: 4)t!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

-(+3/4)t-(-12)=0

We add all the numbers together, and all the variables

-(+3/4)t+12=0

We multiply parentheses

-3t^2+12=0

a = -3; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-3)·12
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-3}=\frac{-12}{-6}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-3}=\frac{12}{-6}
=-2
$