(5x^2-7)=(2x^2+19)

Solution for (5x^2-7)=(2x^2+19) equation:

(5x^2-7)=(2x^2+19)

We move all terms to the left:

(5x^2-7)-((2x^2+19))=0

We get rid of parentheses

5x^2-((2x^2+19))-7=0


We calculate terms in parentheses: -((2x^2+19)), so:

(2x^2+19)

We get rid of parentheses

2x^2+19

Back to the equation:

-(2x^2+19)


We get rid of parentheses

5x^2-2x^2-19-7=0

We add all the numbers together, and all the variables

3x^2-26=0

a = 3; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·3·(-26)
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{78}}{2*3}=\frac{0-2\sqrt{78}}{6}
=-\frac{2\sqrt{78}}{6}
=-\frac{\sqrt{78}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{78}}{2*3}=\frac{0+2\sqrt{78}}{6}
=\frac{2\sqrt{78}}{6}
=\frac{\sqrt{78}}{3}
$