4x^2+28x-119=0.

Solution for 4x^2+28x-119=0. equation:

4x^2+28x-119=0.

a = 4; b = 28; c = -119;
Δ = b2-4ac
Δ = 282-4·4·(-119)
Δ = 2688
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:

$\sqrt{\Delta}=\sqrt{2688}=\sqrt{64*42}=\sqrt{64}*\sqrt{42}=8\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8\sqrt{42}}{2*4}=\frac{-28-8\sqrt{42}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8\sqrt{42}}{2*4}=\frac{-28+8\sqrt{42}}{8}
$