What are the solutions to the equation 4r^2+20r+24=0 ?

The solutions to the equation 4r^2+20r+24=0 are r=-2,r=-3

Find the zeros of 4r^2+20r+24=0

The zeros of 4r^2+20r+24=0 are r=-2,r=-3

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4 r^{2} + 20 r + 24 = 0

r = - 2, r = - 3

Solution steps

4 r^{2} + 20 r + 24 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 24}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 24 = 4

r_{1, 2} = \frac{- 20 \pm 4}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- 20 + 4}{2 \cdot 4}, r_{2} = \frac{- 20 - 4}{2 \cdot 4}

r = \frac{- 20 + 4}{2 \cdot 4} : - 2

r = \frac{- 20 - 4}{2 \cdot 4} : - 3

The solutions to the quadratic equation are:

r = - 2, r = - 3

x (x + 12) = 400

x, x^{2} + 4 x y + y^{2} = 9

2 X^{2} - 6 X = 6 X^{2} - 8 X

4 p^{2} - 12 p + 7 = 0

4 e^{2} + 3 e = - 12

What are the solutions to the equation 4r^2+20r+24=0 ?
The solutions to the equation 4r^2+20r+24=0 are r=-2,r=-3
Find the zeros of 4r^2+20r+24=0
The zeros of 4r^2+20r+24=0 are r=-2,r=-3