M3-5m2+7m-3=0

Solution for M3-5m2+7m-3=0 equation:

3-5M^2+7M-3=0

We add all the numbers together, and all the variables

-5M^2+7M=0

a = -5; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-5)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-5}=\frac{-14}{-10}
=1+2/5
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-5}=\frac{0}{-10}
=0
$