What are the solutions to the equation (x^2)/5+x= 7/5 ?

The solutions to the equation (x^2)/5+x= 7/5 are x=(-5+sqrt(53))/2 ,x=(-5-sqrt(53))/2

Find the zeros of (x^2)/5+x= 7/5

The zeros of (x^2)/5+x= 7/5 are x=(-5+sqrt(53))/2 ,x=(-5-sqrt(53))/2

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\frac{x ^{2}}{5} + x = \frac{7}{5}

x = \frac{- 5 + 53}{2}, x = \frac{- 5 - 53}{2}

Decimal Notation

x = 1.14005 \dots, x = - 6.14005 \dots

Solution steps

\frac{x ^{2}}{5} + x = \frac{7}{5}

Multiply both sides by

5

x^{2} + 5 x = 7

Move

7

to the left side

x^{2} + 5 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 7) = 53

x_{1, 2} = \frac{- 5 \pm 53}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 53}{2 \cdot 1}, x_{2} = \frac{- 5 - 53}{2 \cdot 1}

x = \frac{- 5 + 53}{2 \cdot 1} : \frac{- 5 + 53}{2}

x = \frac{- 5 - 53}{2 \cdot 1} : \frac{- 5 - 53}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 53}{2}, x = \frac{- 5 - 53}{2}

∣ 2 x - 1 ∣ = 4

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

x - 6 \leq 3

2 (- 2) - 5 y = 10

- 5 (2 b + 7) + b < - b - 11

What are the solutions to the equation (x^2)/5+x= 7/5 ?
The solutions to the equation (x^2)/5+x= 7/5 are x=(-5+sqrt(53))/2 ,x=(-5-sqrt(53))/2
Find the zeros of (x^2)/5+x= 7/5
The zeros of (x^2)/5+x= 7/5 are x=(-5+sqrt(53))/2 ,x=(-5-sqrt(53))/2