M(t)=-t^2+4t+5

Solution for M(t)=-t^2+4t+5 equation:

(M)=-M^2+4M+5

We move all terms to the left:

(M)-(-M^2+4M+5)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

M^2-3M-5=0

a = 1; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·1·(-5)
Δ = 29
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{-(-3)-\sqrt{29}}{2*1}=\frac{3-\sqrt{29}}{2}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{29}}{2*1}=\frac{3+\sqrt{29}}{2}
$