(4x^2)+36x-319=0

Solution for (4x^2)+36x-319=0 equation:

(4x^2)+36x-319=0

a = 4; b = 36; c = -319;
Δ = b2-4ac
Δ = 362-4·4·(-319)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-80}{2*4}=\frac{-116}{8}
=-14+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+80}{2*4}=\frac{44}{8}
=5+1/2
$