(2/5)m+1=7

Solution for (2/5)m+1=7 equation:

(2/5)m+1=7

We move all terms to the left:

(2/5)m+1-(7)=0


Domain of the equation: 5)m!=0

m!=0/1

m!=0

m∈R


We add all the numbers together, and all the variables

(+2/5)m+1-7=0

We add all the numbers together, and all the variables

(+2/5)m-6=0

We multiply parentheses

2m^2-6=0

a = 2; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·2·(-6)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*2}=\frac{0-4\sqrt{3}}{4}
=-\frac{4\sqrt{3}}{4}
=-\sqrt{3}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*2}=\frac{0+4\sqrt{3}}{4}
=\frac{4\sqrt{3}}{4}
=\sqrt{3}
$