5t^2+98-490=0

Solution for 5t^2+98-490=0 equation:

5t^2+98-490=0

We add all the numbers together, and all the variables

5t^2-392=0

a = 5; b = 0; c = -392;
Δ = b2-4ac
Δ = 02-4·5·(-392)
Δ = 7840
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:

$\sqrt{\Delta}=\sqrt{7840}=\sqrt{784*10}=\sqrt{784}*\sqrt{10}=28\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{10}}{2*5}=\frac{0-28\sqrt{10}}{10}
=-\frac{28\sqrt{10}}{10}
=-\frac{14\sqrt{10}}{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{10}}{2*5}=\frac{0+28\sqrt{10}}{10}
=\frac{28\sqrt{10}}{10}
=\frac{14\sqrt{10}}{5}
$