4x^2+13=19

Solution for 4x^2+13=19 equation:

4x^2+13=19

We move all terms to the left:

4x^2+13-(19)=0

We add all the numbers together, and all the variables

4x^2-6=0

a = 4; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·4·(-6)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*4}=\frac{0-4\sqrt{6}}{8}
=-\frac{4\sqrt{6}}{8}
=-\frac{\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*4}=\frac{0+4\sqrt{6}}{8}
=\frac{4\sqrt{6}}{8}
=\frac{\sqrt{6}}{2}
$