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

The solutions to the equation 4w^2+w=3 are w= 3/4 ,w=-1

Find the zeros of 4w^2+w=3

The zeros of 4w^2+w=3 are w= 3/4 ,w=-1

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4 w^{2} + w = 3

w = \frac{3}{4}, w = - 1

Decimal Notation

w = 0.75, w = - 1

Solution steps

4 w^{2} + w = 3

Move

3

to the left side

4 w^{2} + w - 3 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

1^{2} - 4 \cdot 4 (- 3) = 7

w_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 4}

Separate the solutions

w_{1} = \frac{- 1 + 7}{2 \cdot 4}, w_{2} = \frac{- 1 - 7}{2 \cdot 4}

w = \frac{- 1 + 7}{2 \cdot 4} : \frac{3}{4}

w = \frac{- 1 - 7}{2 \cdot 4} : - 1

The solutions to the quadratic equation are:

w = \frac{3}{4}, w = - 1

(\frac{5}{4} \times \frac{3}{5}) - (\frac{7}{3} - \frac{11}{7})

9 x + 6

(2 a^{3})^{4} (a^{3})^{3}

x^{2} + 9 x y + 18 y^{2}

u + 5 > 7

What are the solutions to the equation 4w^2+w=3 ?
The solutions to the equation 4w^2+w=3 are w= 3/4 ,w=-1
Find the zeros of 4w^2+w=3
The zeros of 4w^2+w=3 are w= 3/4 ,w=-1