x^2/20+13=40

Solution for x^2/20+13=40 equation:

x^2/20+13=40

We move all terms to the left:

x^2/20+13-(40)=0

We add all the numbers together, and all the variables

x^2/20-27=0

We multiply all the terms by the denominator


x^2-27*20=0

We add all the numbers together, and all the variables

x^2-540=0

a = 1; b = 0; c = -540;
Δ = b2-4ac
Δ = 02-4·1·(-540)
Δ = 2160
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:

$\sqrt{\Delta}=\sqrt{2160}=\sqrt{144*15}=\sqrt{144}*\sqrt{15}=12\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{15}}{2*1}=\frac{0-12\sqrt{15}}{2}
=-\frac{12\sqrt{15}}{2}
=-6\sqrt{15}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{15}}{2*1}=\frac{0+12\sqrt{15}}{2}
=\frac{12\sqrt{15}}{2}
=6\sqrt{15}
$