(5x^2+32)=(7x^2)

Solution for (5x^2+32)=(7x^2) equation:

(5x^2+32)=(7x^2)

We move all terms to the left:

(5x^2+32)-((7x^2))=0

determiningTheFunctionDomain
(5x^2+32)-7x^2=0

We add all the numbers together, and all the variables

-7x^2+(5x^2+32)=0

We get rid of parentheses

-7x^2+5x^2+32=0

We add all the numbers together, and all the variables

-2x^2+32=0

a = -2; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-2)·32
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-2}=\frac{-16}{-4}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-2}=\frac{16}{-4}
=-4
$