2/5x^2=42

Solution for 2/5x^2=42 equation:

2/5x^2=42

We move all terms to the left:

2/5x^2-(42)=0


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

x^2!=0/5

x^2!=√0

x!=0

x∈R


We multiply all the terms by the denominator


-42*5x^2+2=0

Wy multiply elements

-210x^2+2=0

a = -210; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-210)·2
Δ = 1680
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:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{105}}{2*-210}=\frac{0-4\sqrt{105}}{-420}
=-\frac{4\sqrt{105}}{-420}
=-\frac{\sqrt{105}}{-105}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{105}}{2*-210}=\frac{0+4\sqrt{105}}{-420}
=\frac{4\sqrt{105}}{-420}
=\frac{\sqrt{105}}{-105}
$