3x^2+2x-9=(x+3)

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

3x^2+2x-9=(x+3)

We move all terms to the left:

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


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

(x+3)

We get rid of parentheses

x+3

Back to the equation:

-(x+3)


We get rid of parentheses

3x^2+2x-x-3-9=0

We add all the numbers together, and all the variables

3x^2+x-12=0

a = 3; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·3·(-12)
Δ = 145
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{-(1)-\sqrt{145}}{2*3}=\frac{-1-\sqrt{145}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{145}}{2*3}=\frac{-1+\sqrt{145}}{6}
$