3x^2+41=65

Solution for 3x^2+41=65 equation:

3x^2+41=65

We move all terms to the left:

3x^2+41-(65)=0

We add all the numbers together, and all the variables

3x^2-24=0

a = 3; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·3·(-24)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*3}=\frac{0-12\sqrt{2}}{6}
=-\frac{12\sqrt{2}}{6}
=-2\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*3}=\frac{0+12\sqrt{2}}{6}
=\frac{12\sqrt{2}}{6}
=2\sqrt{2}
$