4x^2+64x+119=0

Solution for 4x^2+64x+119=0 equation:

4x^2+64x+119=0

a = 4; b = 64; c = +119;
Δ = b2-4ac
Δ = 642-4·4·119
Δ = 2192
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:

$\sqrt{\Delta}=\sqrt{2192}=\sqrt{16*137}=\sqrt{16}*\sqrt{137}=4\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-4\sqrt{137}}{2*4}=\frac{-64-4\sqrt{137}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+4\sqrt{137}}{2*4}=\frac{-64+4\sqrt{137}}{8}
$