4x^2+24x=16

Solution for 4x^2+24x=16 equation:

4x^2+24x=16

We move all terms to the left:

4x^2+24x-(16)=0

a = 4; b = 24; c = -16;
Δ = b2-4ac
Δ = 242-4·4·(-16)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{13}}{2*4}=\frac{-24-8\sqrt{13}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{13}}{2*4}=\frac{-24+8\sqrt{13}}{8}
$