(3x+4)(x+2)=280

Solution for (3x+4)(x+2)=280 equation:

(3x+4)(x+2)=280

We move all terms to the left:

(3x+4)(x+2)-(280)=0

We multiply parentheses ..

(+3x^2+6x+4x+8)-280=0

We get rid of parentheses

3x^2+6x+4x+8-280=0

We add all the numbers together, and all the variables

3x^2+10x-272=0

a = 3; b = 10; c = -272;
Δ = b2-4ac
Δ = 102-4·3·(-272)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-58}{2*3}=\frac{-68}{6}
=-11+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+58}{2*3}=\frac{48}{6}
=8
$