4x^2=4x+19

Solution for 4x^2=4x+19 equation:

4x^2=4x+19

We move all terms to the left:

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

We get rid of parentheses

4x^2-4x-19=0

a = 4; b = -4; c = -19;
Δ = b2-4ac
Δ = -42-4·4·(-19)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{5}}{2*4}=\frac{4-8\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{5}}{2*4}=\frac{4+8\sqrt{5}}{8}
$