What are the solutions to the equation 4x^2+19x-22=0 ?

The solutions to the equation 4x^2+19x-22=0 are x=(-19+sqrt(713))/8 ,x=(-19-sqrt(713))/8

Find the zeros of 4x^2+19x-22=0

The zeros of 4x^2+19x-22=0 are x=(-19+sqrt(713))/8 ,x=(-19-sqrt(713))/8

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4 x^{2} + 19 x - 22 = 0

x = \frac{- 19 + 713}{8}, x = \frac{- 19 - 713}{8}

Decimal Notation

x = 0.96275 \dots, x = - 5.71275 \dots

Solution steps

4 x^{2} + 19 x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 4 ( - 22 )}{2 \cdot 4}

1 9^{2} - 4 \cdot 4 (- 22) = 713

x_{1, 2} = \frac{- 19 \pm 713}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 19 + 713}{2 \cdot 4}, x_{2} = \frac{- 19 - 713}{2 \cdot 4}

x = \frac{- 19 + 713}{2 \cdot 4} : \frac{- 19 + 713}{8}

x = \frac{- 19 - 713}{2 \cdot 4} : \frac{- 19 - 713}{8}

The solutions to the quadratic equation are:

x = \frac{- 19 + 713}{8}, x = \frac{- 19 - 713}{8}

\frac{( 4 + 4 i )}{( 5 + 4 i )}

- \frac{2}{3} - 4

4 (7 - h) = 8

- 6 x - x + 10 = 15 - 7 x - 5

\frac{t}{4} > 7

What are the solutions to the equation 4x^2+19x-22=0 ?
The solutions to the equation 4x^2+19x-22=0 are x=(-19+sqrt(713))/8 ,x=(-19-sqrt(713))/8
Find the zeros of 4x^2+19x-22=0
The zeros of 4x^2+19x-22=0 are x=(-19+sqrt(713))/8 ,x=(-19-sqrt(713))/8