M(t)=t^2=10t

Solution for M(t)=t^2=10t equation:

(M)=M^2=10M

We move all terms to the left:

(M)-(M^2)=0

determiningTheFunctionDomain
-M^2+M=0

We add all the numbers together, and all the variables

-1M^2+M=0

a = -1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-1)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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