(1/2)(6h-4)=-5h+1

Solution for (1/2)(6h-4)=-5h+1 equation:

(1/2)(6h-4)=-5h+1

We move all terms to the left:

(1/2)(6h-4)-(-5h+1)=0


Domain of the equation: 2)(6h-4)!=0

h∈R


We add all the numbers together, and all the variables

(+1/2)(6h-4)-(-5h+1)=0

We get rid of parentheses

(+1/2)(6h-4)+5h-1=0

We multiply parentheses ..

(+6h^2+1/2*-4)+5h-1=0

We multiply all the terms by the denominator


(+6h^2+1+5h*2*-4)-1*2*-4)=0

We add all the numbers together, and all the variables

(+6h^2+1+5h*2*-4)=0

We get rid of parentheses

6h^2+5h*2*+1-4=0

We add all the numbers together, and all the variables

6h^2+5h*2*-3=0

Wy multiply elements

6h^2+10h^2-3=0

We add all the numbers together, and all the variables

16h^2-3=0

a = 16; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·16·(-3)
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*16}=\frac{0-8\sqrt{3}}{32}
=-\frac{8\sqrt{3}}{32}
=-\frac{\sqrt{3}}{4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*16}=\frac{0+8\sqrt{3}}{32}
=\frac{8\sqrt{3}}{32}
=\frac{\sqrt{3}}{4}
$