5x(x+4)=7+(x-2)

Solution for 5x(x+4)=7+(x-2) equation:

5x(x+4)=7+(x-2)

We move all terms to the left:

5x(x+4)-(7+(x-2))=0

We multiply parentheses

5x^2+20x-(7+(x-2))=0


We calculate terms in parentheses: -(7+(x-2)), so:

7+(x-2)

determiningTheFunctionDomain
(x-2)+7

We get rid of parentheses

x-2+7

We add all the numbers together, and all the variables

x+5

Back to the equation:

-(x+5)


We get rid of parentheses

5x^2+20x-x-5=0

We add all the numbers together, and all the variables

5x^2+19x-5=0

a = 5; b = 19; c = -5;
Δ = b2-4ac
Δ = 192-4·5·(-5)
Δ = 461
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{461}}{2*5}=\frac{-19-\sqrt{461}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{461}}{2*5}=\frac{-19+\sqrt{461}}{10}
$