(1/4)(162+x)=55

Solution for (1/4)(162+x)=55 equation:

(1/4)(162+x)=55

We move all terms to the left:

(1/4)(162+x)-(55)=0


Domain of the equation: 4)(162+x)!=0

x∈R


We add all the numbers together, and all the variables

(+1/4)(x+162)-55=0

We multiply parentheses ..

(+x^2+1/4*162)-55=0

We multiply all the terms by the denominator


(+x^2+1-55*4*162)=0

We get rid of parentheses

x^2+1-55*4*162=0

We add all the numbers together, and all the variables

x^2-35639=0

a = 1; b = 0; c = -35639;
Δ = b2-4ac
Δ = 02-4·1·(-35639)
Δ = 142556
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{142556}=\sqrt{4*35639}=\sqrt{4}*\sqrt{35639}=2\sqrt{35639}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{35639}}{2*1}=\frac{0-2\sqrt{35639}}{2}
=-\frac{2\sqrt{35639}}{2}
=-\sqrt{35639}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{35639}}{2*1}=\frac{0+2\sqrt{35639}}{2}
=\frac{2\sqrt{35639}}{2}
=\sqrt{35639}
$