x^2+10x-19=3

Solution for x^2+10x-19=3 equation:

x^2+10x-19=3

We move all terms to the left:

x^2+10x-19-(3)=0

We add all the numbers together, and all the variables

x^2+10x-22=0

a = 1; b = 10; c = -22;
Δ = b2-4ac
Δ = 102-4·1·(-22)
Δ = 188
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:

$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{47}}{2*1}=\frac{-10-2\sqrt{47}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{47}}{2*1}=\frac{-10+2\sqrt{47}}{2}
$