x^2+14x=48=0

Solution for x^2+14x=48=0 equation:

x^2+14x=48=0

We move all terms to the left:

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

a = 1; b = 14; c = -48;
Δ = b2-4ac
Δ = 142-4·1·(-48)
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{97}}{2*1}=\frac{-14-2\sqrt{97}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{97}}{2*1}=\frac{-14+2\sqrt{97}}{2}
$