4x^2+33=199

Solution for 4x^2+33=199 equation:

4x^2+33=199

We move all terms to the left:

4x^2+33-(199)=0

We add all the numbers together, and all the variables

4x^2-166=0

a = 4; b = 0; c = -166;
Δ = b2-4ac
Δ = 02-4·4·(-166)
Δ = 2656
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:

$\sqrt{\Delta}=\sqrt{2656}=\sqrt{16*166}=\sqrt{16}*\sqrt{166}=4\sqrt{166}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{166}}{2*4}=\frac{0-4\sqrt{166}}{8}
=-\frac{4\sqrt{166}}{8}
=-\frac{\sqrt{166}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{166}}{2*4}=\frac{0+4\sqrt{166}}{8}
=\frac{4\sqrt{166}}{8}
=\frac{\sqrt{166}}{2}
$