(m+3)(m+3)=17

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

(m+3)(m+3)=17

We move all terms to the left:

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

We multiply parentheses ..

(+m^2+3m+3m+9)-17=0

We get rid of parentheses

m^2+3m+3m+9-17=0

We add all the numbers together, and all the variables

m^2+6m-8=0

a = 1; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·1·(-8)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{17}}{2*1}=\frac{-6-2\sqrt{17}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{17}}{2*1}=\frac{-6+2\sqrt{17}}{2}
$