3x^2+5=299

Solution for 3x^2+5=299 equation:

3x^2+5=299

We move all terms to the left:

3x^2+5-(299)=0

We add all the numbers together, and all the variables

3x^2-294=0

a = 3; b = 0; c = -294;
Δ = b2-4ac
Δ = 02-4·3·(-294)
Δ = 3528
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:

$\sqrt{\Delta}=\sqrt{3528}=\sqrt{1764*2}=\sqrt{1764}*\sqrt{2}=42\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42\sqrt{2}}{2*3}=\frac{0-42\sqrt{2}}{6}
=-\frac{42\sqrt{2}}{6}
=-7\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42\sqrt{2}}{2*3}=\frac{0+42\sqrt{2}}{6}
=\frac{42\sqrt{2}}{6}
=7\sqrt{2}
$