4x^2+24x-5=16

Solution for 4x^2+24x-5=16 equation:

4x^2+24x-5=16

We move all terms to the left:

4x^2+24x-5-(16)=0

We add all the numbers together, and all the variables

4x^2+24x-21=0

a = 4; b = 24; c = -21;
Δ = b2-4ac
Δ = 242-4·4·(-21)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{57}}{2*4}=\frac{-24-4\sqrt{57}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{57}}{2*4}=\frac{-24+4\sqrt{57}}{8}
$