16+20x^2=752

Solution for 16+20x^2=752 equation:

16+20x^2=752

We move all terms to the left:

16+20x^2-(752)=0

We add all the numbers together, and all the variables

20x^2-736=0

a = 20; b = 0; c = -736;
Δ = b2-4ac
Δ = 02-4·20·(-736)
Δ = 58880
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:

$\sqrt{\Delta}=\sqrt{58880}=\sqrt{256*230}=\sqrt{256}*\sqrt{230}=16\sqrt{230}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{230}}{2*20}=\frac{0-16\sqrt{230}}{40}
=-\frac{16\sqrt{230}}{40}
=-\frac{2\sqrt{230}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{230}}{2*20}=\frac{0+16\sqrt{230}}{40}
=\frac{16\sqrt{230}}{40}
=\frac{2\sqrt{230}}{5}
$