2x^2+4=19

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

2x^2+4=19

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2-15=0

a = 2; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·2·(-15)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*2}=\frac{0-2\sqrt{30}}{4}
=-\frac{2\sqrt{30}}{4}
=-\frac{\sqrt{30}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*2}=\frac{0+2\sqrt{30}}{4}
=\frac{2\sqrt{30}}{4}
=\frac{\sqrt{30}}{2}
$