x^2+14x+48=400

Solution for x^2+14x+48=400 equation:

x^2+14x+48=400

We move all terms to the left:

x^2+14x+48-(400)=0

We add all the numbers together, and all the variables

x^2+14x-352=0

a = 1; b = 14; c = -352;
Δ = b2-4ac
Δ = 142-4·1·(-352)
Δ = 1604
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:

$\sqrt{\Delta}=\sqrt{1604}=\sqrt{4*401}=\sqrt{4}*\sqrt{401}=2\sqrt{401}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{401}}{2*1}=\frac{-14-2\sqrt{401}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{401}}{2*1}=\frac{-14+2\sqrt{401}}{2}
$