1/5x^2+10=240

Solution for 1/5x^2+10=240 equation:

1/5x^2+10=240

We move all terms to the left:

1/5x^2+10-(240)=0


Domain of the equation: 5x^2!=0

x^2!=0/5

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

1/5x^2-230=0

We multiply all the terms by the denominator


-230*5x^2+1=0

Wy multiply elements

-1150x^2+1=0

a = -1150; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-1150)·1
Δ = 4600
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:

$\sqrt{\Delta}=\sqrt{4600}=\sqrt{100*46}=\sqrt{100}*\sqrt{46}=10\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{46}}{2*-1150}=\frac{0-10\sqrt{46}}{-2300}
=-\frac{10\sqrt{46}}{-2300}
=-\frac{\sqrt{46}}{-230}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{46}}{2*-1150}=\frac{0+10\sqrt{46}}{-2300}
=\frac{10\sqrt{46}}{-2300}
=\frac{\sqrt{46}}{-230}
$