-490t^2+1290t=810

Solution for -490t^2+1290t=810 equation:

-490t^2+1290t=810

We move all terms to the left:

-490t^2+1290t-(810)=0

a = -490; b = 1290; c = -810;
Δ = b2-4ac
Δ = 12902-4·(-490)·(-810)
Δ = 76500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{76500}=\sqrt{900*85}=\sqrt{900}*\sqrt{85}=30\sqrt{85}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1290)-30\sqrt{85}}{2*-490}=\frac{-1290-30\sqrt{85}}{-980}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1290)+30\sqrt{85}}{2*-490}=\frac{-1290+30\sqrt{85}}{-980}
$