(2x^2-1)=25

Solution for (2x^2-1)=25 equation:

(2x^2-1)=25

We move all terms to the left:

(2x^2-1)-(25)=0

We get rid of parentheses

2x^2-1-25=0

We add all the numbers together, and all the variables

2x^2-26=0

a = 2; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·2·(-26)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{13}}{2*2}=\frac{0-4\sqrt{13}}{4}
=-\frac{4\sqrt{13}}{4}
=-\sqrt{13}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{13}}{2*2}=\frac{0+4\sqrt{13}}{4}
=\frac{4\sqrt{13}}{4}
=\sqrt{13}
$