49b^2+48=-98b

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

49b^2+48=-98b

We move all terms to the left:

49b^2+48-(-98b)=0

We get rid of parentheses

49b^2+98b+48=0

a = 49; b = 98; c = +48;
Δ = b2-4ac
Δ = 982-4·49·48
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(98)-14}{2*49}=\frac{-112}{98}
=-1+1/7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(98)+14}{2*49}=\frac{-84}{98}
=-6/7
$