What are the solutions to the equation x^2+5/2 x-5=0 ?

The solutions to the equation x^2+5/2 x-5=0 are x=(-5+sqrt(105))/4 ,x=(-5-sqrt(105))/4

Find the zeros of x^2+5/2 x-5=0

The zeros of x^2+5/2 x-5=0 are x=(-5+sqrt(105))/4 ,x=(-5-sqrt(105))/4

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x^{2} + \frac{5}{2} x - 5 = 0

x = \frac{- 5 + 105}{4}, x = \frac{- 5 - 105}{4}

Decimal

x = 1.31173 \dots, x = - 3.81173 \dots

Solution steps

x^{2} + \frac{5}{2} x - 5 = 0

Multiply both sides by

2

2 x^{2} + 5 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 10 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 10) = 105

x_{1, 2} = \frac{- 5 \pm 105}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 105}{2 \cdot 2}, x_{2} = \frac{- 5 - 105}{2 \cdot 2}

x = \frac{- 5 + 105}{2 \cdot 2} : \frac{- 5 + 105}{4}

x = \frac{- 5 - 105}{2 \cdot 2} : \frac{- 5 - 105}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 105}{4}, x = \frac{- 5 - 105}{4}

x, 2 x - x^{2} = 0.5

3 x^{2} - 9 a x^{2} - x + 3 a

(x - 3)^{2} + (x + 4)^{2} = 29

x^{2} y^{2} + x y = 6, y

a^{2} + n^{2}

What are the solutions to the equation x^2+5/2 x-5=0 ?
The solutions to the equation x^2+5/2 x-5=0 are x=(-5+sqrt(105))/4 ,x=(-5-sqrt(105))/4
Find the zeros of x^2+5/2 x-5=0
The zeros of x^2+5/2 x-5=0 are x=(-5+sqrt(105))/4 ,x=(-5-sqrt(105))/4