(2x+3)(x+4)=67

Solution for (2x+3)(x+4)=67 equation:

(2x+3)(x+4)=67

We move all terms to the left:

(2x+3)(x+4)-(67)=0

We multiply parentheses ..

(+2x^2+8x+3x+12)-67=0

We get rid of parentheses

2x^2+8x+3x+12-67=0

We add all the numbers together, and all the variables

2x^2+11x-55=0

a = 2; b = 11; c = -55;
Δ = b2-4ac
Δ = 112-4·2·(-55)
Δ = 561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{561}}{2*2}=\frac{-11-\sqrt{561}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{561}}{2*2}=\frac{-11+\sqrt{561}}{4}
$