2x+7=19x^2

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

2x+7=19x^2

We move all terms to the left:

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

determiningTheFunctionDomain
-19x^2+2x+7=0

a = -19; b = 2; c = +7;
Δ = b2-4ac
Δ = 22-4·(-19)·7
Δ = 536
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:

$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{134}}{2*-19}=\frac{-2-2\sqrt{134}}{-38}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{134}}{2*-19}=\frac{-2+2\sqrt{134}}{-38}
$