(x+3)(x+4)=18

Solution for (x+3)(x+4)=18 equation:

(x+3)(x+4)=18

We move all terms to the left:

(x+3)(x+4)-(18)=0

We multiply parentheses ..

(+x^2+4x+3x+12)-18=0

We get rid of parentheses

x^2+4x+3x+12-18=0

We add all the numbers together, and all the variables

x^2+7x-6=0

a = 1; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·1·(-6)
Δ = 73
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{-(7)-\sqrt{73}}{2*1}=\frac{-7-\sqrt{73}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{73}}{2*1}=\frac{-7+\sqrt{73}}{2}
$