3(x+5)-5=7+(3+3x2)

Solution for 3(x+5)-5=7+(3+3x2) equation:

3(x+5)-5=7+(3+3x^2)

We move all terms to the left:

3(x+5)-5-(7+(3+3x^2))=0

We multiply parentheses

-(7+(3+3x^2))+3x+15-5=0


We calculate terms in parentheses: -(7+(3+3x^2)), so:

7+(3+3x^2)

determiningTheFunctionDomain
(3+3x^2)+7

We get rid of parentheses

3x^2+3+7

We add all the numbers together, and all the variables

3x^2+10

Back to the equation:

-(3x^2+10)


We add all the numbers together, and all the variables

3x-(3x^2+10)+10=0

We get rid of parentheses

-3x^2+3x-10+10=0

We add all the numbers together, and all the variables

-3x^2+3x=0

a = -3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-3)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-3}=\frac{-6}{-6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-3}=\frac{0}{-6}
=0
$