5x^2+2(2x^2-7)-6=3x^2+28

Solution for 5x^2+2(2x^2-7)-6=3x^2+28 equation:

5x^2+2(2x^2-7)-6=3x^2+28

We move all terms to the left:

5x^2+2(2x^2-7)-6-(3x^2+28)=0

We multiply parentheses

5x^2+4x^2-(3x^2+28)-14-6=0

We get rid of parentheses

5x^2+4x^2-3x^2-28-14-6=0

We add all the numbers together, and all the variables

6x^2-48=0

a = 6; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·6·(-48)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{2}}{2*6}=\frac{0-24\sqrt{2}}{12}
=-\frac{24\sqrt{2}}{12}
=-2\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{2}}{2*6}=\frac{0+24\sqrt{2}}{12}
=\frac{24\sqrt{2}}{12}
=2\sqrt{2}
$