49x^2-28x=5

Solution for 49x^2-28x=5 equation:

49x^2-28x=5

We move all terms to the left:

49x^2-28x-(5)=0

a = 49; b = -28; c = -5;
Δ = b2-4ac
Δ = -282-4·49·(-5)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-42}{2*49}=\frac{-14}{98}
=-1/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+42}{2*49}=\frac{70}{98}
=5/7
$