How do you use the quadratic formula is used to find the roots of the equation $x^2-6x-19=0$ ?

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Answer 1 · 2017-05-03T11:21:55

$x_1=3-2sqrt 7$

$x_2=3+2sqrt 7$

$x^2-6x-19=0$

$"let us find out discriminant of the function "x^2-6x-19$

$Delta=sqrt(b^2-4ac)$

$a=1" , "b=-6" , "c=-19$

$Delta=sqrt((-6)^2-4*1*(-19)$

$Delta=sqrt(36+76)$

$Delta=sqrt(112)$

$Delta=+-4sqrt7$

$x_1=(-b-Delta)/(2a)=(6-4sqrt 7)/(2*1)=(6-4sqrt 7)/2=3-2sqrt 7$

$x_2=(-b+Delta)/(2a)=(6+4sqrt 7)/(2*1)=(6+4sqrt 7)/2=3+2sqrt 7$

Answer 2 · 2017-05-03T11:29:53

Solution : $x=3+2sqrt7 , x= 3- 2sqrt7$

Comparing with general equation $ax^2+bx+c=0$

$x^2-6x-19=0 ; a=1 , b= -6 , c=-19$ . Discriminant $D=b^2-4*a*c = 36+76=112$ is positive so roots are real.

Quadratic formula for finding roots is $x= -b/(2a) +- sqrt(b^2-4ac)/(2a) or x = -(-6)/2+-sqrt112/2 or x = 3 +- 4*sqrt7/2 or x = 3+- 2*sqrt7$

Solution : $x=3+2sqrt7 , x= 3-2sqrt7$ [Ans]

How do you use the quadratic formula is used to find the roots of the equation x^2-6x-19=0x^2-6x-19=0?