49x^2-28x+196=205

Solution for 49x^2-28x+196=205 equation:

49x^2-28x+196=205

We move all terms to the left:

49x^2-28x+196-(205)=0

We add all the numbers together, and all the variables

49x^2-28x-9=0

a = 49; b = -28; c = -9;
Δ = b2-4ac
Δ = -282-4·49·(-9)
Δ = 2548
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2548}=\sqrt{196*13}=\sqrt{196}*\sqrt{13}=14\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-14\sqrt{13}}{2*49}=\frac{28-14\sqrt{13}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+14\sqrt{13}}{2*49}=\frac{28+14\sqrt{13}}{98}
$