4.3t-5.1t^2=0

Solution for 4.3t-5.1t^2=0 equation:

4.3t-5.1t^2=0

a = -5.1; b = 4.3; c = 0;
Δ = b2-4ac
Δ = 4.32-4·(-5.1)·0
Δ = 18.49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4.3)-\sqrt{18.49}}{2*-5.1}=\frac{-4.3-\sqrt{18.49}}{-10.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.3)+\sqrt{18.49}}{2*-5.1}=\frac{-4.3+\sqrt{18.49}}{-10.2}
$