2r^2+1=7/3

Solution for 2r^2+1=7/3 equation:

2r^2+1=7/3

We move all terms to the left:

2r^2+1-(7/3)=0

We add all the numbers together, and all the variables

2r^2+1-(+7/3)=0

We get rid of parentheses

2r^2+1-7/3=0

We multiply all the terms by the denominator


2r^2*3-7+1*3=0

We add all the numbers together, and all the variables

2r^2*3-4=0

Wy multiply elements

6r^2-4=0

a = 6; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·6·(-4)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*6}=\frac{0-4\sqrt{6}}{12}
=-\frac{4\sqrt{6}}{12}
=-\frac{\sqrt{6}}{3}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*6}=\frac{0+4\sqrt{6}}{12}
=\frac{4\sqrt{6}}{12}
=\frac{\sqrt{6}}{3}
$