2/3x^2+5=47

Solution for 2/3x^2+5=47 equation:

2/3x^2+5=47

We move all terms to the left:

2/3x^2+5-(47)=0


Domain of the equation: 3x^2!=0

x^2!=0/3

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

2/3x^2-42=0

We multiply all the terms by the denominator


-42*3x^2+2=0

Wy multiply elements

-126x^2+2=0

a = -126; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-126)·2
Δ = 1008
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:

$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{7}}{2*-126}=\frac{0-12\sqrt{7}}{-252}
=-\frac{12\sqrt{7}}{-252}
=-\frac{\sqrt{7}}{-21}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{7}}{2*-126}=\frac{0+12\sqrt{7}}{-252}
=\frac{12\sqrt{7}}{-252}
=\frac{\sqrt{7}}{-21}
$