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 = 3.67944 \dots, x = - 0.67944 \dots

Solution steps

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

Multiply both sides by

10

10 x^{2} - 30 x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 10 ( - 25 )}{2 \cdot 10}

(- 30)^{2} - 4 \cdot 10 (- 25) = 1019

x_{1, 2} = \frac{- ( - 30 ) \pm 10 19}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 30 ) + 10 19}{2 \cdot 10}, x_{2} = \frac{- ( - 30 ) - 10 19}{2 \cdot 10}

x = \frac{- ( - 30 ) + 10 19}{2 \cdot 10} : \frac{3 + 19}{2}

x = \frac{- ( - 30 ) - 10 19}{2 \cdot 10} : \frac{3 - 19}{2}

The solutions to the quadratic equation are:

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

3 p (g + 5) - 2 (g + 5)

r - 11 > - 17

(\frac{16}{3})^{3} - 8 (\frac{16}{3})^{2}

(x + y)^{3} (3 x - 2 y)^{4} + 2 (x + y)^{4} (3 x - 2 y)^{3}

3 ∣ x - 1 ∣ + 2 \leq 8

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