(3x+1)(x+3)=19-2(x+2)2

Solution for (3x+1)(x+3)=19-2(x+2)2 equation:

(3x+1)(x+3)=19-2(x+2)2

We move all terms to the left:

(3x+1)(x+3)-(19-2(x+2)2)=0

We multiply parentheses ..

(+3x^2+9x+x+3)-(19-2(x+2)2)=0


We calculate terms in parentheses: -(19-2(x+2)2), so:

19-2(x+2)2

determiningTheFunctionDomain
-2(x+2)2+19

We multiply parentheses

-4x-8+19

We add all the numbers together, and all the variables

-4x+11

Back to the equation:

-(-4x+11)


We get rid of parentheses

3x^2+9x+x+4x+3-11=0

We add all the numbers together, and all the variables

3x^2+14x-8=0

a = 3; b = 14; c = -8;
Δ = b2-4ac
Δ = 142-4·3·(-8)
Δ = 292
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{73}}{2*3}=\frac{-14-2\sqrt{73}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{73}}{2*3}=\frac{-14+2\sqrt{73}}{6}
$