(2m-1)(m+2)=m2+16

Solution for (2m-1)(m+2)=m2+16 equation:

(2m-1)(m+2)=m2+16

We move all terms to the left:

(2m-1)(m+2)-(m2+16)=0

We add all the numbers together, and all the variables

-(+m^2+16)+(2m-1)(m+2)=0

We get rid of parentheses

-m^2+(2m-1)(m+2)-16=0

We multiply parentheses ..

-m^2+(+2m^2+4m-1m-2)-16=0

We add all the numbers together, and all the variables

-1m^2+(+2m^2+4m-1m-2)-16=0

We get rid of parentheses

-1m^2+2m^2+4m-1m-2-16=0

We add all the numbers together, and all the variables

m^2+3m-18=0

a = 1; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*1}=\frac{-12}{2}
=-6
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*1}=\frac{6}{2}
=3
$