5=+19x2

Solution for 5=+19x2 equation:

5=+19x^2

We move all terms to the left:

5-(+19x^2)=0

We get rid of parentheses

-19x^2+5=0

a = -19; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-19)·5
Δ = 380
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:

$\sqrt{\Delta}=\sqrt{380}=\sqrt{4*95}=\sqrt{4}*\sqrt{95}=2\sqrt{95}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{95}}{2*-19}=\frac{0-2\sqrt{95}}{-38}
=-\frac{2\sqrt{95}}{-38}
=-\frac{\sqrt{95}}{-19}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{95}}{2*-19}=\frac{0+2\sqrt{95}}{-38}
=\frac{2\sqrt{95}}{-38}
=\frac{\sqrt{95}}{-19}
$