What are the solutions to the equation (x^2+3x-2.5)=0 ?

The solutions to the equation (x^2+3x-2.5)=0 are x=(-3+sqrt(19))/2 ,x=(-3-sqrt(19))/2

Find the zeros of (x^2+3x-2.5)=0

The zeros of (x^2+3x-2.5)=0 are x=(-3+sqrt(19))/2 ,x=(-3-sqrt(19))/2

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(x^{2} + 3 x - 2.5) = 0

x = \frac{- 3 + 19}{2}, x = \frac{- 3 - 19}{2}

Decimal Notation

x = 0.67944 \dots, x = - 3.67944 \dots

Solution steps

(x^{2} + 3 x - 2.5) = 0

Refine

x^{2} + 3 x - 2.5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 2.5 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 2.5) = 19

x_{1, 2} = \frac{- 3 \pm 19}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 19}{2 \cdot 1}, x_{2} = \frac{- 3 - 19}{2 \cdot 1}

x = \frac{- 3 + 19}{2 \cdot 1} : \frac{- 3 + 19}{2}

x = \frac{- 3 - 19}{2 \cdot 1} : \frac{- 3 - 19}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 19}{2}, x = \frac{- 3 - 19}{2}

a^{2} + 4^{2} = 8^{2}

x^{2} - 18 x = - 81

x^{2} - 60 = 4

(\frac{3 y}{4})^{2} + y^{2} = 225

3 y^{2} + 14 y = 5

What are the solutions to the equation (x^2+3x-2.5)=0 ?
The solutions to the equation (x^2+3x-2.5)=0 are x=(-3+sqrt(19))/2 ,x=(-3-sqrt(19))/2
Find the zeros of (x^2+3x-2.5)=0
The zeros of (x^2+3x-2.5)=0 are x=(-3+sqrt(19))/2 ,x=(-3-sqrt(19))/2