What are the solutions to the equation 4x^2+8x-4.5=0 ?

The solutions to the equation 4x^2+8x-4.5=0 are x=(-4+sqrt(34))/4 ,x=-(4+sqrt(34))/4

Find the zeros of 4x^2+8x-4.5=0

The zeros of 4x^2+8x-4.5=0 are x=(-4+sqrt(34))/4 ,x=-(4+sqrt(34))/4

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4 x^{2} + 8 x - 4.5 = 0

x = \frac{- 4 + 34}{4}, x = - \frac{4 + 34}{4}

Decimal Notation

x = 0.45773 \dots, x = - 2.45773 \dots

Solution steps

4 x^{2} + 8 x - 4.5 = 0

Multiply both sides by

10

40 x^{2} + 80 x - 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 40 ( - 45 )}{2 \cdot 40}

8 0^{2} - 4 \cdot 40 (- 45) = 2034

x_{1, 2} = \frac{- 80 \pm 20 34}{2 \cdot 40}

Separate the solutions

x_{1} = \frac{- 80 + 20 34}{2 \cdot 40}, x_{2} = \frac{- 80 - 20 34}{2 \cdot 40}

x = \frac{- 80 + 20 34}{2 \cdot 40} : \frac{- 4 + 34}{4}

x = \frac{- 80 - 20 34}{2 \cdot 40} : - \frac{4 + 34}{4}

The solutions to the quadratic equation are:

x = \frac{- 4 + 34}{4}, x = - \frac{4 + 34}{4}

3 x^{2} - 84 x = - 117

f, f (x + 1) = 5 x^{2} + 3 x - 2

t, d = v t - \frac{1}{2} a t^{2}

40 x^{2} - 13 x + 1 = 0

25 (5 x - 3)^{2} + 6 = 22

What are the solutions to the equation 4x^2+8x-4.5=0 ?
The solutions to the equation 4x^2+8x-4.5=0 are x=(-4+sqrt(34))/4 ,x=-(4+sqrt(34))/4
Find the zeros of 4x^2+8x-4.5=0
The zeros of 4x^2+8x-4.5=0 are x=(-4+sqrt(34))/4 ,x=-(4+sqrt(34))/4