2^2+21=3x^2-4

Solution for 2^2+21=3x^2-4 equation:

2^2+21=3x^2-4

We move all terms to the left:

2^2+21-(3x^2-4)=0

We add all the numbers together, and all the variables

-(3x^2-4)+25=0

We get rid of parentheses

-3x^2+4+25=0

We add all the numbers together, and all the variables

-3x^2+29=0

a = -3; b = 0; c = +29;
Δ = b2-4ac
Δ = 02-4·(-3)·29
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{87}}{2*-3}=\frac{0-2\sqrt{87}}{-6}
=-\frac{2\sqrt{87}}{-6}
=-\frac{\sqrt{87}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{87}}{2*-3}=\frac{0+2\sqrt{87}}{-6}
=\frac{2\sqrt{87}}{-6}
=\frac{\sqrt{87}}{-3}
$