4t^2=196

Solution for 4t^2=196 equation:

4t^2=196

We move all terms to the left:

4t^2-(196)=0

a = 4; b = 0; c = -196;
Δ = b2-4ac
Δ = 02-4·4·(-196)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-56}{2*4}=\frac{-56}{8}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+56}{2*4}=\frac{56}{8}
=7
$