(x+3)(x-4)=48

Solution for (x+3)(x-4)=48 equation:

(x+3)(x-4)=48

We move all terms to the left:

(x+3)(x-4)-(48)=0

We multiply parentheses ..

(+x^2-4x+3x-12)-48=0

We get rid of parentheses

x^2-4x+3x-12-48=0

We add all the numbers together, and all the variables

x^2-1x-60=0

a = 1; b = -1; c = -60;
Δ = b2-4ac
Δ = -12-4·1·(-60)
Δ = 241
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{-(-1)-\sqrt{241}}{2*1}=\frac{1-\sqrt{241}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{241}}{2*1}=\frac{1+\sqrt{241}}{2}
$