What are the solutions to the equation 4w^2+11w+3=-1 ?

The solutions to the equation 4w^2+11w+3=-1 are w=(-11+sqrt(57))/8 ,w=(-11-sqrt(57))/8

Find the zeros of 4w^2+11w+3=-1

The zeros of 4w^2+11w+3=-1 are w=(-11+sqrt(57))/8 ,w=(-11-sqrt(57))/8

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4 w^{2} + 11 w + 3 = - 1

w = \frac{- 11 + 57}{8}, w = \frac{- 11 - 57}{8}

Decimal Notation

w = - 0.43127 \dots, w = - 2.31872 \dots

Solution steps

4 w^{2} + 11 w + 3 = - 1

Move

1

to the left side

4 w^{2} + 11 w + 4 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

1 1^{2} - 4 \cdot 4 \cdot 4 = 57

w_{1, 2} = \frac{- 11 \pm 57}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- 11 + 57}{2 \cdot 4}, w_{2} = \frac{- 11 - 57}{2 \cdot 4}

w = \frac{- 11 + 57}{2 \cdot 4} : \frac{- 11 + 57}{8}

w = \frac{- 11 - 57}{2 \cdot 4} : \frac{- 11 - 57}{8}

The solutions to the quadratic equation are:

w = \frac{- 11 + 57}{8}, w = \frac{- 11 - 57}{8}

5 r^{2} = - 10 + 27 r

u^{2} + 2 u - 8 = 0

x, x^{2} + x y = 12

4 a^{2} + 9 a = 9

x, f (x + 4) = 3 x^{2} - 6 x

What are the solutions to the equation 4w^2+11w+3=-1 ?
The solutions to the equation 4w^2+11w+3=-1 are w=(-11+sqrt(57))/8 ,w=(-11-sqrt(57))/8
Find the zeros of 4w^2+11w+3=-1
The zeros of 4w^2+11w+3=-1 are w=(-11+sqrt(57))/8 ,w=(-11-sqrt(57))/8