(2x+3)(x+4)=228

Solution for (2x+3)(x+4)=228 equation:

(2x+3)(x+4)=228

We move all terms to the left:

(2x+3)(x+4)-(228)=0

We multiply parentheses ..

(+2x^2+8x+3x+12)-228=0

We get rid of parentheses

2x^2+8x+3x+12-228=0

We add all the numbers together, and all the variables

2x^2+11x-216=0

a = 2; b = 11; c = -216;
Δ = b2-4ac
Δ = 112-4·2·(-216)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-43}{2*2}=\frac{-54}{4}
=-13+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+43}{2*2}=\frac{32}{4}
=8
$