5t2+15t-490=0

Solution for 5t2+15t-490=0 equation:

5t^2+15t-490=0

a = 5; b = 15; c = -490;
Δ = b2-4ac
Δ = 152-4·5·(-490)
Δ = 10025
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:

$\sqrt{\Delta}=\sqrt{10025}=\sqrt{25*401}=\sqrt{25}*\sqrt{401}=5\sqrt{401}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{401}}{2*5}=\frac{-15-5\sqrt{401}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{401}}{2*5}=\frac{-15+5\sqrt{401}}{10}
$