(3x+1)(3x+1)=(x+9)(x+4)-42

Solution for (3x+1)(3x+1)=(x+9)(x+4)-42 equation:

(3x+1)(3x+1)=(x+9)(x+4)-42

We move all terms to the left:

(3x+1)(3x+1)-((x+9)(x+4)-42)=0

We multiply parentheses ..

(+9x^2+3x+3x+1)-((x+9)(x+4)-42)=0


We calculate terms in parentheses: -((x+9)(x+4)-42), so:

(x+9)(x+4)-42

We multiply parentheses ..

(+x^2+4x+9x+36)-42

We get rid of parentheses

x^2+4x+9x+36-42

We add all the numbers together, and all the variables

x^2+13x-6

Back to the equation:

-(x^2+13x-6)


We get rid of parentheses

9x^2-x^2+3x+3x-13x+1+6=0

We add all the numbers together, and all the variables

8x^2-7x+7=0

a = 8; b = -7; c = +7;
Δ = b2-4ac
Δ = -72-4·8·7
Δ = -175
Delta is less than zero, so there is no solution for the equation