1t^2-4t=0

Solution for 1t^2-4t=0 equation:

1t^2-4t=0

We add all the numbers together, and all the variables

t^2-4t=0

a = 1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·1·0
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*1}=\frac{0}{2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*1}=\frac{8}{2}
=4
$