(X-5)(x+4)=10

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Solution for (X-5)(x+4)=10 equation:



(X-5)(X+4)=10
We move all terms to the left:
(X-5)(X+4)-(10)=0
We multiply parentheses ..
(+X^2+4X-5X-20)-10=0
We get rid of parentheses
X^2+4X-5X-20-10=0
We add all the numbers together, and all the variables
X^2-1X-30=0
a = 1; b = -1; c = -30;
Δ = b2-4ac
Δ = -12-4·1·(-30)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*1}=\frac{-10}{2} =-5 $
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*1}=\frac{12}{2} =6 $

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