49t^2+140t=-98

Solution for 49t^2+140t=-98 equation:

49t^2+140t=-98

We move all terms to the left:

49t^2+140t-(-98)=0

We add all the numbers together, and all the variables

49t^2+140t+98=0

a = 49; b = 140; c = +98;
Δ = b2-4ac
Δ = 1402-4·49·98
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(140)-14\sqrt{2}}{2*49}=\frac{-140-14\sqrt{2}}{98}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(140)+14\sqrt{2}}{2*49}=\frac{-140+14\sqrt{2}}{98}
$