# I need help on this equation: x(x+4)=21

kjcdb8er | Teacher | (Level 1) Associate Educator

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Another way to solve quadratic equations is to use the quadratic formula. Any equation of the form ax² + bx + c = 0 can be solved for x by the formula:
x = (-b² - 4ac))/2a
So so solve your problem x(x+4) =  21, first expand it into the form above:
x(x + 4) - 21 = 0          Subtract 21 from both sides
x² + 4x - 21 = 0           Now the equation is in the form ax² + bx + c = 0

x = (-4² - 4*1*(-21)))/2*1
x =  ( -4 ± sqrt(16 + 4*21))/2
x =  ( -4 ± sqrt(16 + 84))/2
x =  ( -4 ± sqrt(100))/2
x =  ( -4 ± 10)/2 = -2 ± 5
So x = -2 + 5 = 3    and   x = -2 - 5 = -7

jess1999 | Student, Grade 9 | (Level 1) Valedictorian

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x ( x + 4 ) = 21

To solve this equation , we should first distribute the " x "

By distributing the " x " , you should get

x^2 + 4x = 21 now we can subtract 21 on both sides

By subtracting 21 on both sides , you would get

x^2 + 4x - 21 = 0 Now we can factor ( such as using the x factor which means having two numbers that would have  a product of -21 and a sum of 4 )

By factoring the equation , you would get

( x + 7 ) ( x - 3 ) This is just basically x + 7 = 0 and x - 3 = 0

mrmdkunkel | Middle School Teacher | eNotes Newbie

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x = 3

I noticed when looking at this equation that it was easier to solve than others were making it.  I looked at the factors of 21 (3 and 7) and paired them each with the other side of the equation (x=3 and (x+4)=7). To me it was as simple as that.

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

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x(x+4)=21,, x^2+4x-21=0,, (x+7)(x-3)=0 so (x+7)=0 or (x-3)=0 so X=3 or x=(-7)

neela | High School Teacher | (Level 3) Valedictorian

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x(x+4)=21.

To solve for x, we expand left side.

x^2+4x=21.

The left side is made a perfect square by adding 4. So right side also is added 4, to maintain the equality  as below:

x^2+4x+4=21+4

(x+2)^2=25.

We take square root:

x+2=+5   or   x+2 =-5

x=5-2= 3 or x= -5-2=-7.

This can also be done by factorisation as below:

x(x+4)=21

Expand the left and shift the 21 to left by simple operation of subtraction. Then we are  with aquadratic  equation in x.

x^2+4x=21.

x^2+4x-21=0.

x^2+7x-3x-21=0, as the middle term 4x=7x-3x and (7x)(-3x)= -21x^2 , the product of first and last terms:x^2*(-21)

x(x+7)-3(x+7)=0

(x+7)(x-3)=0---->x+7=0   or x-3=0

x+7=0------> x = -7

x-3=0------> x = 3