7x^2+1=13

Solution for 7x^2+1=13 equation:

7x^2+1=13

We move all terms to the left:

7x^2+1-(13)=0

We add all the numbers together, and all the variables

7x^2-12=0

a = 7; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·7·(-12)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*7}=\frac{0-4\sqrt{21}}{14}
=-\frac{4\sqrt{21}}{14}
=-\frac{2\sqrt{21}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*7}=\frac{0+4\sqrt{21}}{14}
=\frac{4\sqrt{21}}{14}
=\frac{2\sqrt{21}}{7}
$