3b*b-28b+49=0

Solution for 3b*b-28b+49=0 equation:

3b*b-28b+49=0

We add all the numbers together, and all the variables

-28b+3b*b+49=0

Wy multiply elements

3b^2-28b+49=0

a = 3; b = -28; c = +49;
Δ = b2-4ac
Δ = -282-4·3·49
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{-(-28)-14}{2*3}=\frac{14}{6}
=2+1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+14}{2*3}=\frac{42}{6}
=7
$