1/4t+4=3)4(t+8

Solution for 1/4t+4=3)4(t+8 equation:

1/4t+4=3)4(t+8

We move all terms to the left:

1/4t+4-(3)4(t+8)=0


Domain of the equation: 4t!=0

t!=0/4

t!=0

t∈R


We multiply parentheses

1/4t-34t-272+4=0

We multiply all the terms by the denominator


-34t*4t-272*4t+4*4t+1=0

Wy multiply elements

-136t^2-1088t+16t+1=0

We add all the numbers together, and all the variables

-136t^2-1072t+1=0

a = -136; b = -1072; c = +1;
Δ = b2-4ac
Δ = -10722-4·(-136)·1
Δ = 1149728
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:

$\sqrt{\Delta}=\sqrt{1149728}=\sqrt{16*71858}=\sqrt{16}*\sqrt{71858}=4\sqrt{71858}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1072)-4\sqrt{71858}}{2*-136}=\frac{1072-4\sqrt{71858}}{-272}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1072)+4\sqrt{71858}}{2*-136}=\frac{1072+4\sqrt{71858}}{-272}
$