3(x2+4)+5=-6(x2+2x)+13

Solution for 3(x2+4)+5=-6(x2+2x)+13 equation:

3(x2+4)+5=-6(x2+2x)+13

We move all terms to the left:

3(x2+4)+5-(-6(x2+2x)+13)=0

We add all the numbers together, and all the variables

3(+x^2+4)-(-6(+x^2+2x)+13)+5=0

We multiply parentheses

3x^2-(-6(+x^2+2x)+13)+12+5=0


We calculate terms in parentheses: -(-6(+x^2+2x)+13), so:

-6(+x^2+2x)+13

We multiply parentheses

-6x^2-12x+13

Back to the equation:

-(-6x^2-12x+13)


We add all the numbers together, and all the variables

3x^2-(-6x^2-12x+13)+17=0

We get rid of parentheses

3x^2+6x^2+12x-13+17=0

We add all the numbers together, and all the variables

9x^2+12x+4=0

a = 9; b = 12; c = +4;
Δ = b2-4ac
Δ = 122-4·9·4
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-12}{18}=-2/3$