7x^2=0+19

Solution for 7x^2=0+19 equation:

7x^2=0+19

We move all terms to the left:

7x^2-(0+19)=0

We add all the numbers together, and all the variables

7x^2-19=0

a = 7; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·7·(-19)
Δ = 532
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:

$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{133}}{2*7}=\frac{0-2\sqrt{133}}{14}
=-\frac{2\sqrt{133}}{14}
=-\frac{\sqrt{133}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{133}}{2*7}=\frac{0+2\sqrt{133}}{14}
=\frac{2\sqrt{133}}{14}
=\frac{\sqrt{133}}{7}
$