(m^2+3)/4=21

Solution for (m^2+3)/4=21 equation:

(m^2+3)/4=21

We move all terms to the left:

(m^2+3)/4-(21)=0

We multiply all the terms by the denominator


(m^2+3)-21*4=0

We add all the numbers together, and all the variables

(m^2+3)-84=0

We get rid of parentheses

m^2+3-84=0

We add all the numbers together, and all the variables

m^2-81=0

a = 1; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·1·(-81)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*1}=\frac{-18}{2}
=-9
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*1}=\frac{18}{2}
=9
$