4.1(m+3)m=1.5

Solution for 4.1(m+3)m=1.5 equation:

4.1(m+3)m=1.5

We move all terms to the left:

4.1(m+3)m-(1.5)=0

We add all the numbers together, and all the variables

4.1(m+3)m-1.5=0

We multiply parentheses

4m^2+12m-1.5=0

a = 4; b = 12; c = -1.5;
Δ = b2-4ac
Δ = 122-4·4·(-1.5)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{42}}{2*4}=\frac{-12-2\sqrt{42}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{42}}{2*4}=\frac{-12+2\sqrt{42}}{8}
$