(4x^2-19)=(2x^2-17)

Solution for (4x^2-19)=(2x^2-17) equation:

(4x^2-19)=(2x^2-17)

We move all terms to the left:

(4x^2-19)-((2x^2-17))=0

We get rid of parentheses

4x^2-((2x^2-17))-19=0


We calculate terms in parentheses: -((2x^2-17)), so:

(2x^2-17)

We get rid of parentheses

2x^2-17

Back to the equation:

-(2x^2-17)


We get rid of parentheses

4x^2-2x^2+17-19=0

We add all the numbers together, and all the variables

2x^2-2=0

a = 2; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·2·(-2)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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