(m-5)(m+3)=17

Solution for (m-5)(m+3)=17 equation:

(m-5)(m+3)=17

We move all terms to the left:

(m-5)(m+3)-(17)=0

We multiply parentheses ..

(+m^2+3m-5m-15)-17=0

We get rid of parentheses

m^2+3m-5m-15-17=0

We add all the numbers together, and all the variables

m^2-2m-32=0

a = 1; b = -2; c = -32;
Δ = b2-4ac
Δ = -22-4·1·(-32)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{33}}{2*1}=\frac{2-2\sqrt{33}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{33}}{2*1}=\frac{2+2\sqrt{33}}{2}
$