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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+m^2-5m+2m-10)-3=0

We get rid of parentheses

m^2-5m+2m-10-3=0

We add all the numbers together, and all the variables

m^2-3m-13=0

a = 1; b = -3; c = -13;
Δ = b2-4ac
Δ = -32-4·1·(-13)
Δ = 61
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{-(-3)-\sqrt{61}}{2*1}=\frac{3-\sqrt{61}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{61}}{2*1}=\frac{3+\sqrt{61}}{2}
$