49x^2-21+2=0

Solution for 49x^2-21+2=0 equation:

49x^2-21+2=0

We add all the numbers together, and all the variables

49x^2-19=0

a = 49; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·49·(-19)
Δ = 3724
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:

$\sqrt{\Delta}=\sqrt{3724}=\sqrt{196*19}=\sqrt{196}*\sqrt{19}=14\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{19}}{2*49}=\frac{0-14\sqrt{19}}{98}
=-\frac{14\sqrt{19}}{98}
=-\frac{\sqrt{19}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{19}}{2*49}=\frac{0+14\sqrt{19}}{98}
=\frac{14\sqrt{19}}{98}
=\frac{\sqrt{19}}{7}
$