15m(m+5)=32m-20

Solution for 15m(m+5)=32m-20 equation:

15m(m+5)=32m-20

We move all terms to the left:

15m(m+5)-(32m-20)=0

We multiply parentheses

15m^2+75m-(32m-20)=0

We get rid of parentheses

15m^2+75m-32m+20=0

We add all the numbers together, and all the variables

15m^2+43m+20=0

a = 15; b = 43; c = +20;
Δ = b2-4ac
Δ = 432-4·15·20
Δ = 649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-\sqrt{649}}{2*15}=\frac{-43-\sqrt{649}}{30}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+\sqrt{649}}{2*15}=\frac{-43+\sqrt{649}}{30}
$