(1/2)(4d-2)=d+5

Solution for (1/2)(4d-2)=d+5 equation:

(1/2)(4d-2)=d+5

We move all terms to the left:

(1/2)(4d-2)-(d+5)=0


Domain of the equation: 2)(4d-2)!=0

d∈R


We add all the numbers together, and all the variables

(+1/2)(4d-2)-(d+5)=0

We get rid of parentheses

(+1/2)(4d-2)-d-5=0

We multiply parentheses ..

(+4d^2+1/2*-2)-d-5=0

We multiply all the terms by the denominator


(+4d^2+1-d*2*-2)-5*2*-2)=0

We add all the numbers together, and all the variables

(+4d^2+1-d*2*-2)=0

We get rid of parentheses

4d^2-d*2*+1-2=0

We add all the numbers together, and all the variables

4d^2-d*2*-1=0

Wy multiply elements

4d^2-2d^2-1=0

We add all the numbers together, and all the variables

2d^2-1=0

a = 2; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·2·(-1)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*2}=\frac{0-2\sqrt{2}}{4}
=-\frac{2\sqrt{2}}{4}
=-\frac{\sqrt{2}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*2}=\frac{0+2\sqrt{2}}{4}
=\frac{2\sqrt{2}}{4}
=\frac{\sqrt{2}}{2}
$