(X-4)(x+10)=14

Solution for (X-4)(x+10)=14 equation:

(X-4)(X+10)=14

We move all terms to the left:

(X-4)(X+10)-(14)=0

We multiply parentheses ..

(+X^2+10X-4X-40)-14=0

We get rid of parentheses

X^2+10X-4X-40-14=0

We add all the numbers together, and all the variables

X^2+6X-54=0

a = 1; b = 6; c = -54;
Δ = b2-4ac
Δ = 62-4·1·(-54)
Δ = 252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{7}}{2*1}=\frac{-6-6\sqrt{7}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{7}}{2*1}=\frac{-6+6\sqrt{7}}{2}
$