(3/4)t+(7/8)t=39

Solution for (3/4)t+(7/8)t=39 equation:

(3/4)t+(7/8)t=39

We move all terms to the left:

(3/4)t+(7/8)t-(39)=0


Domain of the equation: 4)t!=0

t!=0/1

t!=0

t∈R



Domain of the equation: 8)t!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+3/4)t+(+7/8)t-39=0

We multiply parentheses

3t^2+7t^2-39=0

We add all the numbers together, and all the variables

10t^2-39=0

a = 10; b = 0; c = -39;
Δ = b2-4ac
Δ = 02-4·10·(-39)
Δ = 1560
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:

$\sqrt{\Delta}=\sqrt{1560}=\sqrt{4*390}=\sqrt{4}*\sqrt{390}=2\sqrt{390}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{390}}{2*10}=\frac{0-2\sqrt{390}}{20}
=-\frac{2\sqrt{390}}{20}
=-\frac{\sqrt{390}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{390}}{2*10}=\frac{0+2\sqrt{390}}{20}
=\frac{2\sqrt{390}}{20}
=\frac{\sqrt{390}}{10}
$