D=4t+5t^2

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

=4D+5D^2

We move all terms to the left:

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

We get rid of parentheses

-5D^2-4D=0

a = -5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-5)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-5}=\frac{0}{-10}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-5}=\frac{8}{-10}
=-4/5
$