What are the solutions to the equation 4x^2+20x+23=0 ?

The solutions to the equation 4x^2+20x+23=0 are x=(-5+sqrt(2))/2 ,x=-(5+sqrt(2))/2

Find the zeros of 4x^2+20x+23=0

The zeros of 4x^2+20x+23=0 are x=(-5+sqrt(2))/2 ,x=-(5+sqrt(2))/2

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4 x^{2} + 20 x + 23 = 0

x = \frac{- 5 + 2}{2}, x = - \frac{5 + 2}{2}

Decimal Notation

x = - 1.79289 \dots, x = - 3.20710 \dots

Solution steps

4 x^{2} + 20 x + 23 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 23}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 23 = 42

x_{1, 2} = \frac{- 20 \pm 4 2}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 4 2}{2 \cdot 4}, x_{2} = \frac{- 20 - 4 2}{2 \cdot 4}

x = \frac{- 20 + 4 2}{2 \cdot 4} : \frac{- 5 + 2}{2}

x = \frac{- 20 - 4 2}{2 \cdot 4} : - \frac{5 + 2}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 2}{2}, x = - \frac{5 + 2}{2}

x^{2} + 18 x + 42 = 0

3 x^{2} + 10 x + 9 = 0

1 - 5 x + x^{2} + 2 x = 1

(x + 1) (x - 2) = 3

\frac{a + c x ^{2}}{b} = x^{2}

What are the solutions to the equation 4x^2+20x+23=0 ?
The solutions to the equation 4x^2+20x+23=0 are x=(-5+sqrt(2))/2 ,x=-(5+sqrt(2))/2
Find the zeros of 4x^2+20x+23=0
The zeros of 4x^2+20x+23=0 are x=(-5+sqrt(2))/2 ,x=-(5+sqrt(2))/2