What are the solutions to the equation 4u^2+24u+29=1 ?

The solutions to the equation 4u^2+24u+29=1 are u=-3+sqrt(2),u=-3-sqrt(2)

Find the zeros of 4u^2+24u+29=1

The zeros of 4u^2+24u+29=1 are u=-3+sqrt(2),u=-3-sqrt(2)

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4 u^{2} + 24 u + 29 = 1

u = - 3 + 2, u = - 3 - 2

Decimal Notation

u = - 1.58578 \dots, u = - 4.41421 \dots

Solution steps

4 u^{2} + 24 u + 29 = 1

Move

1

to the left side

4 u^{2} + 24 u + 28 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 4 \cdot 28}{2 \cdot 4}

2 4^{2} - 4 \cdot 4 \cdot 28 = 82

u_{1, 2} = \frac{- 24 \pm 8 2}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 24 + 8 2}{2 \cdot 4}, u_{2} = \frac{- 24 - 8 2}{2 \cdot 4}

u = \frac{- 24 + 8 2}{2 \cdot 4} : - 3 + 2

u = \frac{- 24 - 8 2}{2 \cdot 4} : - 3 - 2

The solutions to the quadratic equation are:

u = - 3 + 2, u = - 3 - 2

2 (- 4 s + 2) \geq - 5 s - 10

\frac{k ^{6}}{k ^{6}}

ln (ab)

5 x^{2} = - 7 + 12 x

7 x + 11 x = - 54

What are the solutions to the equation 4u^2+24u+29=1 ?
The solutions to the equation 4u^2+24u+29=1 are u=-3+sqrt(2),u=-3-sqrt(2)
Find the zeros of 4u^2+24u+29=1
The zeros of 4u^2+24u+29=1 are u=-3+sqrt(2),u=-3-sqrt(2)