9x^2+48-122=x^2+6

Solution for 9x^2+48-122=x^2+6 equation:

9x^2+48-122=x^2+6

We move all terms to the left:

9x^2+48-122-(x^2+6)=0

We add all the numbers together, and all the variables

9x^2-(x^2+6)-74=0

We get rid of parentheses

9x^2-x^2-6-74=0

We add all the numbers together, and all the variables

8x^2-80=0

a = 8; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·8·(-80)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*8}=\frac{0-16\sqrt{10}}{16}
=-\frac{16\sqrt{10}}{16}
=-\sqrt{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*8}=\frac{0+16\sqrt{10}}{16}
=\frac{16\sqrt{10}}{16}
=\sqrt{10}
$