4(x^2-9)-(4-x^2-9)=-5

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

4(x^2-9)-(4-x^2-9)=-5

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-(4-x^2-9)+4x^2-36+5=0

We get rid of parentheses

x^2+4x^2-4+9-36+5=0

We add all the numbers together, and all the variables

5x^2-26=0

a = 5; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·5·(-26)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{130}}{2*5}=\frac{0-2\sqrt{130}}{10}
=-\frac{2\sqrt{130}}{10}
=-\frac{\sqrt{130}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{130}}{2*5}=\frac{0+2\sqrt{130}}{10}
=\frac{2\sqrt{130}}{10}
=\frac{\sqrt{130}}{5}
$