((4x+x)x)+x-48=0

Solution for ((4x+x)x)+x-48=0 equation:

((4x+x)x)+x-48=0

We add all the numbers together, and all the variables

((+5x)x)+x-48=0

We add all the numbers together, and all the variables

x+((+5x)x)-48=0


We calculate terms in parentheses: +((+5x)x), so:

(+5x)x

We multiply parentheses

5x^2

Back to the equation:

+(5x^2)


determiningTheFunctionDomain
5x^2+x-48=0

a = 5; b = 1; c = -48;
Δ = b2-4ac
Δ = 12-4·5·(-48)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*5}=\frac{-32}{10}
=-3+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*5}=\frac{30}{10}
=3
$