(x^2+2x+5=)+(3x^2+9x+7=)

Solution for (x^2+2x+5=)+(3x^2+9x+7=) equation:

(x^2+2x+5=)+(3x^2+9x+7=)

We move all terms to the left:

(x^2+2x+5-()+(3x^2+9x+7)=0


We calculate terms in parentheses: +(x^2+2x+5-()+(3x^2+9x+7), so:

x^2+2x+5-()+(3x^2+9x+7

determiningTheFunctionDomain
x^2+2x+(3x^2+9x+7+5-()


We calculate terms in parentheses: +(3x^2+9x+7+5-(), so:

3x^2+9x+7+5-(

We add all the numbers together, and all the variables

3x^2+9x

Back to the equation:

+(3x^2+9x)


We get rid of parentheses

x^2+3x^2+2x+9x

We add all the numbers together, and all the variables

4x^2+11x

Back to the equation:

+(4x^2+11x)


We get rid of parentheses

4x^2+11x=0

a = 4; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·4·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*4}=\frac{-22}{8}
=-2+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*4}=\frac{0}{8}
=0
$