What are the solutions to the equation 4x^2+20x+16=0 ?

The solutions to the equation 4x^2+20x+16=0 are x=-1,x=-4

Find the zeros of 4x^2+20x+16=0

The zeros of 4x^2+20x+16=0 are x=-1,x=-4

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4 x^{2} + 20 x + 16 = 0

x = - 1, x = - 4

Solution steps

4 x^{2} + 20 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 16 = 12

x_{1, 2} = \frac{- 20 \pm 12}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 12}{2 \cdot 4}, x_{2} = \frac{- 20 - 12}{2 \cdot 4}

x = \frac{- 20 + 12}{2 \cdot 4} : - 1

x = \frac{- 20 - 12}{2 \cdot 4} : - 4

The solutions to the quadratic equation are:

x = - 1, x = - 4

t, s = - 16 t^{2} + 96 t

y^{2} + 5 y + 5 = 0

x^{2} - 8 x - 76 = 8

- x^{2} + 100 = 0

s^{2} - 2 s - 8 = 0

What are the solutions to the equation 4x^2+20x+16=0 ?
The solutions to the equation 4x^2+20x+16=0 are x=-1,x=-4
Find the zeros of 4x^2+20x+16=0
The zeros of 4x^2+20x+16=0 are x=-1,x=-4