(x-2)(x+3)=10

Solution for (x-2)(x+3)=10 equation:

(x-2)(x+3)=10

We move all terms to the left:

(x-2)(x+3)-(10)=0

We multiply parentheses ..

(+x^2+3x-2x-6)-10=0

We get rid of parentheses

x^2+3x-2x-6-10=0

We add all the numbers together, and all the variables

x^2+x-16=0

a = 1; b = 1; c = -16;
Δ = b2-4ac
Δ = 12-4·1·(-16)
Δ = 65
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{65}}{2*1}=\frac{-1-\sqrt{65}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*1}=\frac{-1+\sqrt{65}}{2}
$