D=5t+4t^2

Solution for D=5t+4t^2 equation:

=5D+4D^2

We move all terms to the left:

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

We get rid of parentheses

-4D^2-5D=0

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

$\sqrt{\Delta}=\sqrt{25}=5$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-4}=\frac{0}{-8}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-4}=\frac{10}{-8}
=-1+1/4
$