1/6(12t+48)-20=1/8(24t-144)

Solution for 1/6(12t+48)-20=1/8(24t-144) equation:

1/6(12t+48)-20=1/8(24t-144)

We move all terms to the left:

1/6(12t+48)-20-(1/8(24t-144))=0


Domain of the equation: 6(12t+48)!=0

t∈R



Domain of the equation: 8(24t-144))!=0

t∈R


We calculate fractions


(8t2/(6(12t+48)*8(24t-144)))+(-6t1/(6(12t+48)*8(24t-144)))-20=0


We calculate terms in parentheses: +(8t2/(6(12t+48)*8(24t-144))), so:

8t2/(6(12t+48)*8(24t-144))

We multiply all the terms by the denominator


8t2

We add all the numbers together, and all the variables

8t^2

Back to the equation:

+(8t^2)



We calculate terms in parentheses: +(-6t1/(6(12t+48)*8(24t-144))), so:

-6t1/(6(12t+48)*8(24t-144))

We multiply all the terms by the denominator


-6t1

We add all the numbers together, and all the variables

-6t

Back to the equation:

+(-6t)


We get rid of parentheses

8t^2-6t-20=0

a = 8; b = -6; c = -20;
Δ = b2-4ac
Δ = -62-4·8·(-20)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-26}{2*8}=\frac{-20}{16}
=-1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+26}{2*8}=\frac{32}{16}
=2
$