(1m+5)(3m+4)=3(1m+2)-2

Solution for (1m+5)(3m+4)=3(1m+2)-2 equation:

(1m+5)(3m+4)=3(1m+2)-2

We move all terms to the left:

(1m+5)(3m+4)-(3(1m+2)-2)=0

We add all the numbers together, and all the variables

(m+5)(3m+4)-(3(m+2)-2)=0

We multiply parentheses ..

(+3m^2+4m+15m+20)-(3(m+2)-2)=0


We calculate terms in parentheses: -(3(m+2)-2), so:

3(m+2)-2

We multiply parentheses

3m+6-2

We add all the numbers together, and all the variables

3m+4

Back to the equation:

-(3m+4)


We get rid of parentheses

3m^2+4m+15m-3m+20-4=0

We add all the numbers together, and all the variables

3m^2+16m+16=0

a = 3; b = 16; c = +16;
Δ = b2-4ac
Δ = 162-4·3·16
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*3}=\frac{-24}{6}
=-4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*3}=\frac{-8}{6}
=-1+1/3
$