(4m-3)4=-(9+4m)/8

Solution for (4m-3)4=-(9+4m)/8 equation:

(4m-3)4=-(9+4m)/8

We move all terms to the left:

(4m-3)4-(-(9+4m)/8)=0

We add all the numbers together, and all the variables

(4m-3)4-(-(4m+9)/8)=0

We multiply parentheses

16m-(-(4m+9)/8)-12=0

We multiply all the terms by the denominator


16m*8)-(-(4m+9)-12*8)=0

We add all the numbers together, and all the variables

16m*8)-(-(4m+9)=0

Wy multiply elements

128m^2-(4m+9)=0

We get rid of parentheses

128m^2-4m-9=0

a = 128; b = -4; c = -9;
Δ = b2-4ac
Δ = -42-4·128·(-9)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-68}{2*128}=\frac{-64}{256}
=-1/4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+68}{2*128}=\frac{72}{256}
=9/32
$