What are the solutions to the equation 4r^2+20r+25=0 ?

The solutions to the equation 4r^2+20r+25=0 are r=-5/2

Find the zeros of 4r^2+20r+25=0

The zeros of 4r^2+20r+25=0 are r=-5/2

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4 r^{2} + 20 r + 25 = 0

r = - \frac{5}{2}

Solution steps

4 r^{2} + 20 r + 25 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 25}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 25 = 0

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

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

\frac{- 20}{2 \cdot 4} = - \frac{5}{2}

r = - \frac{5}{2}

The solution to the quadratic equation is:

r = - \frac{5}{2}

\frac{2}{5} \cdot 20

\frac{- 9 y ^{2}}{5 x} \cdot \frac{- 10 x ^{2}}{3 y}

x^{2} + x + \frac{1}{4} = 0

x^{3} - 9 x^{2} + 108

20 x^{2} - 7 x - 6 = 0

What are the solutions to the equation 4r^2+20r+25=0 ?
The solutions to the equation 4r^2+20r+25=0 are r=-5/2
Find the zeros of 4r^2+20r+25=0
The zeros of 4r^2+20r+25=0 are r=-5/2