4m^2+2m-19=3m^2

Solution for 4m^2+2m-19=3m^2 equation:

4m^2+2m-19=3m^2

We move all terms to the left:

4m^2+2m-19-(3m^2)=0

determiningTheFunctionDomain
4m^2-3m^2+2m-19=0

We add all the numbers together, and all the variables

m^2+2m-19=0

a = 1; b = 2; c = -19;
Δ = b2-4ac
Δ = 22-4·1·(-19)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{5}}{2*1}=\frac{-2-4\sqrt{5}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{5}}{2*1}=\frac{-2+4\sqrt{5}}{2}
$