What are the solutions to the equation 4r^2=-28r ?

The solutions to the equation 4r^2=-28r are r=0,r=-7

Find the zeros of 4r^2=-28r

The zeros of 4r^2=-28r are r=0,r=-7

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4 r^{2} = - 28 r

r = 0, r = - 7

Solution steps

4 r^{2} = - 28 r

Move

28 r

to the left side

4 r^{2} + 28 r = 0

Factor

4 r^{2} + 28 r : 4 r (r + 7)

4 r (r + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r = 0 or r + 7 = 0

Solve

r + 7 = 0 : r = - 7

The solutions to the quadratic equation are:

r = 0, r = - 7

- 3 - 8

y, x^{2} y + x^{2} - x y^{2} + x y + 2 = 0

lo g_{2} (8) = 3 x

(X + 3)^{2} = 2 X + 6

11 q + 5 \leq 49

What are the solutions to the equation 4r^2=-28r ?
The solutions to the equation 4r^2=-28r are r=0,r=-7
Find the zeros of 4r^2=-28r
The zeros of 4r^2=-28r are r=0,r=-7