8(x^2+5)=2(x^2+26)

Solution for 8(x^2+5)=2(x^2+26) equation:

8(x^2+5)=2(x^2+26)

We move all terms to the left:

8(x^2+5)-(2(x^2+26))=0

We multiply parentheses

8x^2-(2(x^2+26))+40=0


We calculate terms in parentheses: -(2(x^2+26)), so:

2(x^2+26)

We multiply parentheses

2x^2+52

Back to the equation:

-(2x^2+52)


We get rid of parentheses

8x^2-2x^2-52+40=0

We add all the numbers together, and all the variables

6x^2-12=0

a = 6; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·6·(-12)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*6}=\frac{0-12\sqrt{2}}{12}
=-\frac{12\sqrt{2}}{12}
=-\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*6}=\frac{0+12\sqrt{2}}{12}
=\frac{12\sqrt{2}}{12}
=\sqrt{2}
$