M4=(x^2-27)

Solution for M4=(x^2-27) equation:

4=(M^2-27)

We move all terms to the left:

4-((M^2-27))=0


We calculate terms in parentheses: -((M^2-27)), so:

(M^2-27)

We get rid of parentheses

M^2-27

Back to the equation:

-(M^2-27)


We get rid of parentheses

-M^2+27+4=0

We add all the numbers together, and all the variables

-1M^2+31=0

a = -1; b = 0; c = +31;
Δ = b2-4ac
Δ = 02-4·(-1)·31
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{31}}{2*-1}=\frac{0-2\sqrt{31}}{-2}
=-\frac{2\sqrt{31}}{-2}
=-\frac{\sqrt{31}}{-1}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{31}}{2*-1}=\frac{0+2\sqrt{31}}{-2}
=\frac{2\sqrt{31}}{-2}
=\frac{\sqrt{31}}{-1}
$