M^4-3m^2+2m-5=0

Solution for M^4-3m^2+2m-5=0 equation:

^4-3M^2+2M-5=0

We add all the numbers together, and all the variables

-3M^2+2M=0

a = -3; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-3)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-3}=\frac{-4}{-6}
=2/3
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-3}=\frac{0}{-6}
=0
$