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

The solutions to the equation 4u^2+24u+4=0 are u=-3+2sqrt(2),u=-3-2sqrt(2)

Find the zeros of 4u^2+24u+4=0

The zeros of 4u^2+24u+4=0 are u=-3+2sqrt(2),u=-3-2sqrt(2)

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4 u^{2} + 24 u + 4 = 0

u = - 3 + 22, u = - 3 - 22

Decimal Notation

u = - 0.17157 \dots, u = - 5.82842 \dots

Solution steps

4 u^{2} + 24 u + 4 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

2 4^{2} - 4 \cdot 4 \cdot 4 = 162

u_{1, 2} = \frac{- 24 \pm 16 2}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 24 + 16 2}{2 \cdot 4}, u_{2} = \frac{- 24 - 16 2}{2 \cdot 4}

u = \frac{- 24 + 16 2}{2 \cdot 4} : - 3 + 22

u = \frac{- 24 - 16 2}{2 \cdot 4} : - 3 - 22

The solutions to the quadratic equation are:

u = - 3 + 22, u = - 3 - 22

z^{2} - 10 z + 34 = 0

1 - \frac{7}{30} x - \frac{1}{15} x^{2} = 0

(x + 9) (x - 3) = 0

a, 9 a^{2} - a^{2} = c^{2}

x (3 (x - 4)) + (x - 3) + (4 x + 2) + (2 x - 1) = 300

What are the solutions to the equation 4u^2+24u+4=0 ?
The solutions to the equation 4u^2+24u+4=0 are u=-3+2sqrt(2),u=-3-2sqrt(2)
Find the zeros of 4u^2+24u+4=0
The zeros of 4u^2+24u+4=0 are u=-3+2sqrt(2),u=-3-2sqrt(2)