What are the solutions to the equation 4x^2+19x+21=0 ?

The solutions to the equation 4x^2+19x+21=0 are x=-7/4 ,x=-3

Find the zeros of 4x^2+19x+21=0

The zeros of 4x^2+19x+21=0 are x=-7/4 ,x=-3

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4 x^{2} + 19 x + 21 = 0

x = - \frac{7}{4}, x = - 3

Decimal Notation

x = - 1.75, x = - 3

Solution steps

4 x^{2} + 19 x + 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 4 \cdot 21}{2 \cdot 4}

1 9^{2} - 4 \cdot 4 \cdot 21 = 5

x_{1, 2} = \frac{- 19 \pm 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 19 + 5}{2 \cdot 4}, x_{2} = \frac{- 19 - 5}{2 \cdot 4}

x = \frac{- 19 + 5}{2 \cdot 4} : - \frac{7}{4}

x = \frac{- 19 - 5}{2 \cdot 4} : - 3

The solutions to the quadratic equation are:

x = - \frac{7}{4}, x = - 3

4 x^{2} + 28 x = - 45

10 = 4 (\frac{3 + b}{2})

3 x^{3} - 4 x^{2} - 7 x - 2

\frac{- 12 + - 45}{24}

\frac{1042 - 214}{5}

What are the solutions to the equation 4x^2+19x+21=0 ?
The solutions to the equation 4x^2+19x+21=0 are x=-7/4 ,x=-3
Find the zeros of 4x^2+19x+21=0
The zeros of 4x^2+19x+21=0 are x=-7/4 ,x=-3