48b^2-98+49=0

Solution for 48b^2-98+49=0 equation:

48b^2-98+49=0

We add all the numbers together, and all the variables

48b^2-49=0

a = 48; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·48·(-49)
Δ = 9408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9408}=\sqrt{3136*3}=\sqrt{3136}*\sqrt{3}=56\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-56\sqrt{3}}{2*48}=\frac{0-56\sqrt{3}}{96}
=-\frac{56\sqrt{3}}{96}
=-\frac{7\sqrt{3}}{12}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+56\sqrt{3}}{2*48}=\frac{0+56\sqrt{3}}{96}
=\frac{56\sqrt{3}}{96}
=\frac{7\sqrt{3}}{12}
$