-4t^2+28t+0=0

Solution for -4t^2+28t+0=0 equation:

-4t^2+28t+0=0

We add all the numbers together, and all the variables

-4t^2+28t=0

a = -4; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·(-4)·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*-4}=\frac{-56}{-8}
=+7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*-4}=\frac{0}{-8}
=0
$