-490t^2+1470t=t

Solution for -490t^2+1470t=t equation:

-490t^2+1470t=t

We move all terms to the left:

-490t^2+1470t-(t)=0

We add all the numbers together, and all the variables

-490t^2+1469t=0

a = -490; b = 1469; c = 0;
Δ = b2-4ac
Δ = 14692-4·(-490)·0
Δ = 2157961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2157961}=1469$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1469)-1469}{2*-490}=\frac{-2938}{-980}
=2+489/490
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1469)+1469}{2*-490}=\frac{0}{-980}
=0
$