What are the solutions to the equation 5x^2-9x=-2 ?

The solutions to the equation 5x^2-9x=-2 are x=(9+sqrt(41))/(10),x=(9-sqrt(41))/(10)

Find the zeros of 5x^2-9x=-2

The zeros of 5x^2-9x=-2 are x=(9+sqrt(41))/(10),x=(9-sqrt(41))/(10)

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5 x^{2} - 9 x = - 2

x = \frac{9 + 41}{10}, x = \frac{9 - 41}{10}

Decimal Notation

x = 1.54031 \dots, x = 0.25968 \dots

Solution steps

5 x^{2} - 9 x = - 2

Move

2

to the left side

5 x^{2} - 9 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 \cdot 2}{2 \cdot 5}

(- 9)^{2} - 4 \cdot 5 \cdot 2 = 41

x_{1, 2} = \frac{- ( - 9 ) \pm 41}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 41}{2 \cdot 5}, x_{2} = \frac{- ( - 9 ) - 41}{2 \cdot 5}

x = \frac{- ( - 9 ) + 41}{2 \cdot 5} : \frac{9 + 41}{10}

x = \frac{- ( - 9 ) - 41}{2 \cdot 5} : \frac{9 - 41}{10}

The solutions to the quadratic equation are:

x = \frac{9 + 41}{10}, x = \frac{9 - 41}{10}

x^{2} - 7 x - 78

3 x^{2} - 6 x > 0

7 (x + 1) = 4 - 2 (7 x + 5)

\frac{1}{4} - \frac{1}{4}

\frac{4}{5} + \frac{5}{6}

What are the solutions to the equation 5x^2-9x=-2 ?
The solutions to the equation 5x^2-9x=-2 are x=(9+sqrt(41))/(10),x=(9-sqrt(41))/(10)
Find the zeros of 5x^2-9x=-2
The zeros of 5x^2-9x=-2 are x=(9+sqrt(41))/(10),x=(9-sqrt(41))/(10)