4x2+19x+9=x+4

Solution for 4x2+19x+9=x+4 equation:

4x^2+19x+9=x+4

We move all terms to the left:

4x^2+19x+9-(x+4)=0

We get rid of parentheses

4x^2+19x-x-4+9=0

We add all the numbers together, and all the variables

4x^2+18x+5=0

a = 4; b = 18; c = +5;
Δ = b2-4ac
Δ = 182-4·4·5
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{61}}{2*4}=\frac{-18-2\sqrt{61}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{61}}{2*4}=\frac{-18+2\sqrt{61}}{8}
$