(5x^2-27)=(4x^2-6)

Solution for (5x^2-27)=(4x^2-6) equation:

(5x^2-27)=(4x^2-6)

We move all terms to the left:

(5x^2-27)-((4x^2-6))=0

We get rid of parentheses

5x^2-((4x^2-6))-27=0


We calculate terms in parentheses: -((4x^2-6)), so:

(4x^2-6)

We get rid of parentheses

4x^2-6

Back to the equation:

-(4x^2-6)


We get rid of parentheses

5x^2-4x^2+6-27=0

We add all the numbers together, and all the variables

x^2-21=0

a = 1; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·1·(-21)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{21}}{2*1}=\frac{0-2\sqrt{21}}{2}
=-\frac{2\sqrt{21}}{2}
=-\sqrt{21}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{21}}{2*1}=\frac{0+2\sqrt{21}}{2}
=\frac{2\sqrt{21}}{2}
=\sqrt{21}
$