43=43-4t-5t^2

Solution for 43=43-4t-5t^2 equation:

43=43-4t-5t^2

We move all terms to the left:

43-(43-4t-5t^2)=0

We get rid of parentheses

5t^2+4t-43+43=0

We add all the numbers together, and all the variables

5t^2+4t=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:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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