4x^2+28x-320=0

Solution for 4x^2+28x-320=0 equation:

4x^2+28x-320=0

a = 4; b = 28; c = -320;
Δ = b2-4ac
Δ = 282-4·4·(-320)
Δ = 5904
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:

$\sqrt{\Delta}=\sqrt{5904}=\sqrt{144*41}=\sqrt{144}*\sqrt{41}=12\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-12\sqrt{41}}{2*4}=\frac{-28-12\sqrt{41}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+12\sqrt{41}}{2*4}=\frac{-28+12\sqrt{41}}{8}
$