4x^2+22x-122=0

Solution for 4x^2+22x-122=0 equation:

4x^2+22x-122=0

a = 4; b = 22; c = -122;
Δ = b2-4ac
Δ = 222-4·4·(-122)
Δ = 2436
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:

$\sqrt{\Delta}=\sqrt{2436}=\sqrt{4*609}=\sqrt{4}*\sqrt{609}=2\sqrt{609}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{609}}{2*4}=\frac{-22-2\sqrt{609}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{609}}{2*4}=\frac{-22+2\sqrt{609}}{8}
$