-4.9t^2+100t=510

Solution for -4.9t^2+100t=510 equation:

-4.9t^2+100t=510

We move all terms to the left:

-4.9t^2+100t-(510)=0

a = -4.9; b = 100; c = -510;
Δ = b2-4ac
Δ = 1002-4·(-4.9)·(-510)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-2}{2*-4.9}=\frac{-102}{-9.8}
=10+4/9.8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+2}{2*-4.9}=\frac{-98}{-9.8}
=+10
$