3x^2+14=49

Solution for 3x^2+14=49 equation:

3x^2+14=49

We move all terms to the left:

3x^2+14-(49)=0

We add all the numbers together, and all the variables

3x^2-35=0

a = 3; b = 0; c = -35;
Δ = b2-4ac
Δ = 02-4·3·(-35)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{105}}{2*3}=\frac{0-2\sqrt{105}}{6}
=-\frac{2\sqrt{105}}{6}
=-\frac{\sqrt{105}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{105}}{2*3}=\frac{0+2\sqrt{105}}{6}
=\frac{2\sqrt{105}}{6}
=\frac{\sqrt{105}}{3}
$