3x(4+x)=49

Solution for 3x(4+x)=49 equation:

3x(4+x)=49

We move all terms to the left:

3x(4+x)-(49)=0

We add all the numbers together, and all the variables

3x(x+4)-49=0

We multiply parentheses

3x^2+12x-49=0

a = 3; b = 12; c = -49;
Δ = b2-4ac
Δ = 122-4·3·(-49)
Δ = 732
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:

$\sqrt{\Delta}=\sqrt{732}=\sqrt{4*183}=\sqrt{4}*\sqrt{183}=2\sqrt{183}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{183}}{2*3}=\frac{-12-2\sqrt{183}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{183}}{2*3}=\frac{-12+2\sqrt{183}}{6}
$