2m+3=4-m^2

Solution for 2m+3=4-m^2 equation:

2m+3=4-m^2

We move all terms to the left:

2m+3-(4-m^2)=0

We get rid of parentheses

m^2+2m-4+3=0

We add all the numbers together, and all the variables

m^2+2m-1=0

a = 1; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·1·(-1)
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2}}{2*1}=\frac{-2-2\sqrt{2}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2}}{2*1}=\frac{-2+2\sqrt{2}}{2}
$