7x^2+13=99

Solution for 7x^2+13=99 equation:

7x^2+13=99

We move all terms to the left:

7x^2+13-(99)=0

We add all the numbers together, and all the variables

7x^2-86=0

a = 7; b = 0; c = -86;
Δ = b2-4ac
Δ = 02-4·7·(-86)
Δ = 2408
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:

$\sqrt{\Delta}=\sqrt{2408}=\sqrt{4*602}=\sqrt{4}*\sqrt{602}=2\sqrt{602}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{602}}{2*7}=\frac{0-2\sqrt{602}}{14}
=-\frac{2\sqrt{602}}{14}
=-\frac{\sqrt{602}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{602}}{2*7}=\frac{0+2\sqrt{602}}{14}
=\frac{2\sqrt{602}}{14}
=\frac{\sqrt{602}}{7}
$