4m+15=m^2

Solution for 4m+15=m^2 equation:

4m+15=m^2

We move all terms to the left:

4m+15-(m^2)=0

determiningTheFunctionDomain
-m^2+4m+15=0

We add all the numbers together, and all the variables

-1m^2+4m+15=0

a = -1; b = 4; c = +15;
Δ = b2-4ac
Δ = 42-4·(-1)·15
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{19}}{2*-1}=\frac{-4-2\sqrt{19}}{-2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{19}}{2*-1}=\frac{-4+2\sqrt{19}}{-2}
$