# Solve x(x+4)=20

Asked on by oshiwin

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation x(x + 4) = 20 has to be solved for x.

x(x + 4) = 20

=> x^2 + 4x - 20 = 0

=> x1 = `(-4 + sqrt(16 + 80))/2` and x2 = `(-4 - sqrt(16 + 80))/2`

=> x1 = `-2 + sqrt 24` and x2 = `-2 - sqrt 24`

The solution of the equation x(x + 4) = 20 is `-2 + sqrt 24` and `-2 - sqrt 24`

chaobas | College Teacher | (Level 1) Valedictorian

Posted on

We have the equation as

x(x+4)=20

x^2+4x=20

x^2+4x-20=0

It is in the form of ax^2+bx+c=0, which is a quadratic equation aand here

a=1:b=4 and c=-20

andd we know that

x=(-b+sqr(b^2-4ac))/2a or x=(-b-sqr(b^2-4ac))/2a

x=(-4+sqr((-4)^2-4.1.(-20)))/2.1 or x=(-4-sqr((-4)^2-4.1.(-20)))/2.1

x=(-4+sqr(16+80))/2 or x=(-4-sqr(16+80))/2

x=(-4+sqr(96))/2 or x=(-4-sqr(96))/2

x=(-4+4sqr(6))/2 or x=(-4-4sqr(6))/2

x=-4(1-sqr(6))/2 or x=-4(1+sqr(6))/2

so x = 2(1-sqr(6))  or x=2(1+sqr(6))

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