(3/4)m+(1/8)m-6=15

Solution for (3/4)m+(1/8)m-6=15 equation:

(3/4)m+(1/8)m-6=15

We move all terms to the left:

(3/4)m+(1/8)m-6-(15)=0


Domain of the equation: 4)m!=0

m!=0/1

m!=0

m∈R



Domain of the equation: 8)m!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(+3/4)m+(+1/8)m-6-15=0

We add all the numbers together, and all the variables

(+3/4)m+(+1/8)m-21=0

We multiply parentheses

3m^2+m^2-21=0

We add all the numbers together, and all the variables

4m^2-21=0

a = 4; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·4·(-21)
Δ = 336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*4}=\frac{0-4\sqrt{21}}{8}
=-\frac{4\sqrt{21}}{8}
=-\frac{\sqrt{21}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*4}=\frac{0+4\sqrt{21}}{8}
=\frac{4\sqrt{21}}{8}
=\frac{\sqrt{21}}{2}
$