(x^2-3x-40)/(x^2+7x+10)=0

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Solution for (x^2-3x-40)/(x^2+7x+10)=0 equation: 


D( x )

x^2+7*x+10 = 0

x^2+7*x+10 = 0

x^2+7*x+10 = 0

x^2+7*x+10 = 0

DELTA = 7^2-(1*4*10)

DELTA = 9

DELTA > 0

x = (9^(1/2)-7)/(1*2) or x = (-9^(1/2)-7)/(1*2)

x = -2 or x = -5

x in (-oo:-5) U (-5:-2) U (-2:+oo)

(x^2-(3*x)-40)/(x^2+7*x+10) = 0

(x^2-3*x-40)/(x^2+7*x+10) = 0

x^2-3*x-40 = 0

x^2-3*x-40 = 0

DELTA = (-3)^2-(-40*1*4)

DELTA = 169

DELTA > 0

x = (169^(1/2)+3)/(1*2) or x = (3-169^(1/2))/(1*2)

x = 8 or x = -5

(x+5)*(x-8) = 0

x^2+7*x+10 = 0

x^2+7*x+10 = 0

DELTA = 7^2-(1*4*10)

DELTA = 9

DELTA > 0

x = (9^(1/2)-7)/(1*2) or x = (-9^(1/2)-7)/(1*2)

x = -2 or x = -5

(x+5)*(x+2) = 0

((x+5)*(x-8))/((x+5)*(x+2)) = 0

( x+5 )

x+5 = 0 // - 5

x = -5

( x-8 )

x-8 = 0 // + 8

x = 8

x in { -5} 

x = 8

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