8x^2+28x=480

Solution for 8x^2+28x=480 equation:

8x^2+28x=480

We move all terms to the left:

8x^2+28x-(480)=0

a = 8; b = 28; c = -480;
Δ = b2-4ac
Δ = 282-4·8·(-480)
Δ = 16144
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:

$\sqrt{\Delta}=\sqrt{16144}=\sqrt{16*1009}=\sqrt{16}*\sqrt{1009}=4\sqrt{1009}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{1009}}{2*8}=\frac{-28-4\sqrt{1009}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{1009}}{2*8}=\frac{-28+4\sqrt{1009}}{16}
$