7x^2+14x-49=0

Solution for 7x^2+14x-49=0 equation:

7x^2+14x-49=0

a = 7; b = 14; c = -49;
Δ = b2-4ac
Δ = 142-4·7·(-49)
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-28\sqrt{2}}{2*7}=\frac{-14-28\sqrt{2}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+28\sqrt{2}}{2*7}=\frac{-14+28\sqrt{2}}{14}
$