49b^2-28b+4=

Solution for 49b^2-28b+4= equation:

49b^2-28b+4=

We move all terms to the left:

49b^2-28b+4-()=0

We add all the numbers together, and all the variables

49b^2-28b=0

a = 49; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·49·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*49}=\frac{0}{98}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*49}=\frac{56}{98}
=4/7
$