What are the solutions to the equation 16z^2+40z+25=0 ?

The solutions to the equation 16z^2+40z+25=0 are z=-5/4

Find the zeros of 16z^2+40z+25=0

The zeros of 16z^2+40z+25=0 are z=-5/4

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16 z^{2} + 40 z + 25 = 0

z = - \frac{5}{4}

Decimal Notation

z = - 1.25

Solution steps

16 z^{2} + 40 z + 25 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 16 \cdot 25}{2 \cdot 16}

4 0^{2} - 4 \cdot 16 \cdot 25 = 0

z_{1, 2} = \frac{- 40 \pm 0}{2 \cdot 16}

z = \frac{- 40}{2 \cdot 16}

\frac{- 40}{2 \cdot 16} = - \frac{5}{4}

z = - \frac{5}{4}

The solution to the quadratic equation is:

z = - \frac{5}{4}

x^{2} - 8 x + 8 = - 4

x^{2} = 1 - (- \frac{5}{6})^{2}

x^{2} - 2 x - 5 = 30

b^{2} - 9 b + 20 = 0

49 x^{2} = 121

What are the solutions to the equation 16z^2+40z+25=0 ?
The solutions to the equation 16z^2+40z+25=0 are z=-5/4
Find the zeros of 16z^2+40z+25=0
The zeros of 16z^2+40z+25=0 are z=-5/4