4m^2-21m+10=5m^2

Solution for 4m^2-21m+10=5m^2 equation:

4m^2-21m+10=5m^2

We move all terms to the left:

4m^2-21m+10-(5m^2)=0

determiningTheFunctionDomain
4m^2-5m^2-21m+10=0

We add all the numbers together, and all the variables

-1m^2-21m+10=0

a = -1; b = -21; c = +10;
Δ = b2-4ac
Δ = -212-4·(-1)·10
Δ = 481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{481}}{2*-1}=\frac{21-\sqrt{481}}{-2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{481}}{2*-1}=\frac{21+\sqrt{481}}{-2}
$