x^2-14x+49=40

Solution for x^2-14x+49=40 equation:

x^2-14x+49=40

We move all terms to the left:

x^2-14x+49-(40)=0

We add all the numbers together, and all the variables

x^2-14x+9=0

a = 1; b = -14; c = +9;
Δ = b2-4ac
Δ = -142-4·1·9
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4\sqrt{10}}{2*1}=\frac{14-4\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{10}}{2*1}=\frac{14+4\sqrt{10}}{2}
$