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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2m^2+2m-18=0

a = 2; b = 2; c = -18;
Δ = b2-4ac
Δ = 22-4·2·(-18)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{37}}{2*2}=\frac{-2-2\sqrt{37}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{37}}{2*2}=\frac{-2+2\sqrt{37}}{4}
$