2*x^2+4*x-224=0

Solution for 2*x^2+4*x-224=0 equation:

2x^2+4x-224=0

a = 2; b = 4; c = -224;
Δ = b2-4ac
Δ = 42-4·2·(-224)
Δ = 1808
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:

$\sqrt{\Delta}=\sqrt{1808}=\sqrt{16*113}=\sqrt{16}*\sqrt{113}=4\sqrt{113}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{113}}{2*2}=\frac{-4-4\sqrt{113}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{113}}{2*2}=\frac{-4+4\sqrt{113}}{4}
$