(1x/2)+7-(1x/4)=(19/2)

Solution for (1x/2)+7-(1x/4)=(19/2) equation:

(1x/2)+7-(1x/4)=(19/2)

We move all terms to the left:

(1x/2)+7-(1x/4)-((19/2))=0

We add all the numbers together, and all the variables

(+1x/2)-(+1x/4)+7-((+19/2))=0

We get rid of parentheses

1x/2-1x/4+7-((+19/2))=0

We calculate fractions


(-8x^2)/()+4x/()+7+()/()=0

We add all the numbers together, and all the variables

(-8x^2)/()+4x/()+8=0

We multiply all the terms by the denominator


(-8x^2)+4x+8*()=0

We add all the numbers together, and all the variables

(-8x^2)+4x=0

We get rid of parentheses

-8x^2+4x=0

a = -8; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-8)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-8}=\frac{-8}{-16}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-8}=\frac{0}{-16}
=0
$