(2x^2-5-7)-(4^2+9-5)=0

Solution for (2x^2-5-7)-(4^2+9-5)=0 equation:

(2x^2-5-7)-(4^2+9-5)=0

We add all the numbers together, and all the variables

(2x^2-5-7)-20=0

We get rid of parentheses

2x^2-5-7-20=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
$