2x2+19x2=382

Solution for 2x2+19x2=382 equation:

2x^2+19x^2=382

We move all terms to the left:

2x^2+19x^2-(382)=0

We add all the numbers together, and all the variables

21x^2-382=0

a = 21; b = 0; c = -382;
Δ = b2-4ac
Δ = 02-4·21·(-382)
Δ = 32088
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:

$\sqrt{\Delta}=\sqrt{32088}=\sqrt{4*8022}=\sqrt{4}*\sqrt{8022}=2\sqrt{8022}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{8022}}{2*21}=\frac{0-2\sqrt{8022}}{42}
=-\frac{2\sqrt{8022}}{42}
=-\frac{\sqrt{8022}}{21}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{8022}}{2*21}=\frac{0+2\sqrt{8022}}{42}
=\frac{2\sqrt{8022}}{42}
=\frac{\sqrt{8022}}{21}
$