49b^2+42b+8=0

Solution for 49b^2+42b+8=0 equation:

49b^2+42b+8=0

a = 49; b = 42; c = +8;
Δ = b2-4ac
Δ = 422-4·49·8
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{-(42)-14}{2*49}=\frac{-56}{98}
=-4/7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+14}{2*49}=\frac{-28}{98}
=-2/7
$