49x^2+112x+63=1

Solution for 49x^2+112x+63=1 equation:

49x^2+112x+63=1

We move all terms to the left:

49x^2+112x+63-(1)=0

We add all the numbers together, and all the variables

49x^2+112x+62=0

a = 49; b = 112; c = +62;
Δ = b2-4ac
Δ = 1122-4·49·62
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-14\sqrt{2}}{2*49}=\frac{-112-14\sqrt{2}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+14\sqrt{2}}{2*49}=\frac{-112+14\sqrt{2}}{98}
$