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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


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

(4x^2-6x)

We get rid of parentheses

4x^2-6x

Back to the equation:

-(4x^2-6x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+6x-27=0

a = 1; b = 6; c = -27;
Δ = b2-4ac
Δ = 62-4·1·(-27)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12}{2*1}=\frac{6}{2}
=3
$