(x+2)(2x-7)=x^2-4

Solution for (x+2)(2x-7)=x^2-4 equation:

(x+2)(2x-7)=x^2-4

We move all terms to the left:

(x+2)(2x-7)-(x^2-4)=0

We get rid of parentheses

-x^2+(x+2)(2x-7)+4=0

We multiply parentheses ..

-x^2+(+2x^2-7x+4x-14)+4=0

We add all the numbers together, and all the variables

-1x^2+(+2x^2-7x+4x-14)+4=0

We get rid of parentheses

-1x^2+2x^2-7x+4x-14+4=0

We add all the numbers together, and all the variables

x^2-3x-10=0

a = 1; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*1}=\frac{10}{2}
=5
$