8x^2+41=329

Solution for 8x^2+41=329 equation:

8x^2+41=329

We move all terms to the left:

8x^2+41-(329)=0

We add all the numbers together, and all the variables

8x^2-288=0

a = 8; b = 0; c = -288;
Δ = b2-4ac
Δ = 02-4·8·(-288)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-96}{2*8}=\frac{-96}{16}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+96}{2*8}=\frac{96}{16}
=6
$