2x^2+14x-50=0

Solution for 2x^2+14x-50=0 equation:

2x^2+14x-50=0

a = 2; b = 14; c = -50;
Δ = b2-4ac
Δ = 142-4·2·(-50)
Δ = 596
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:

$\sqrt{\Delta}=\sqrt{596}=\sqrt{4*149}=\sqrt{4}*\sqrt{149}=2\sqrt{149}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{149}}{2*2}=\frac{-14-2\sqrt{149}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{149}}{2*2}=\frac{-14+2\sqrt{149}}{4}
$