10x2-19=-6

Solution for 10x2-19=-6 equation:

10x^2-19=-6

We move all terms to the left:

10x^2-19-(-6)=0

We add all the numbers together, and all the variables

10x^2-13=0

a = 10; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·10·(-13)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{130}}{2*10}=\frac{0-2\sqrt{130}}{20}
=-\frac{2\sqrt{130}}{20}
=-\frac{\sqrt{130}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{130}}{2*10}=\frac{0+2\sqrt{130}}{20}
=\frac{2\sqrt{130}}{20}
=\frac{\sqrt{130}}{10}
$