M(t)=-t2+6t

Solution for M(t)=-t2+6t equation:

(M)=-M2+6M

We move all terms to the left:

(M)-(-M2+6M)=0

We add all the numbers together, and all the variables

-(-1M^2+6M)+M=0

We get rid of parentheses

1M^2-6M+M=0

We add all the numbers together, and all the variables

M^2-5M=0

a = 1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·1·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*1}=\frac{0}{2}
=0
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*1}=\frac{10}{2}
=5
$