(x+2)(3x+4)=300

Solution for (x+2)(3x+4)=300 equation:

(x+2)(3x+4)=300

We move all terms to the left:

(x+2)(3x+4)-(300)=0

We multiply parentheses ..

(+3x^2+4x+6x+8)-300=0

We get rid of parentheses

3x^2+4x+6x+8-300=0

We add all the numbers together, and all the variables

3x^2+10x-292=0

a = 3; b = 10; c = -292;
Δ = b2-4ac
Δ = 102-4·3·(-292)
Δ = 3604
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:

$\sqrt{\Delta}=\sqrt{3604}=\sqrt{4*901}=\sqrt{4}*\sqrt{901}=2\sqrt{901}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{901}}{2*3}=\frac{-10-2\sqrt{901}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{901}}{2*3}=\frac{-10+2\sqrt{901}}{6}
$