What are the solutions to the equation x^2-4/5 x=3 ?

The solutions to the equation x^2-4/5 x=3 are x=(2+sqrt(79))/5 ,x=(2-sqrt(79))/5

Find the zeros of x^2-4/5 x=3

The zeros of x^2-4/5 x=3 are x=(2+sqrt(79))/5 ,x=(2-sqrt(79))/5

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x^{2} - \frac{4}{5} x = 3

x = \frac{2 + 79}{5}, x = \frac{2 - 79}{5}

Decimal Notation

x = 2.17763 \dots, x = - 1.37763 \dots

Solution steps

x^{2} - \frac{4}{5} x = 3

Multiply both sides by

5

5 x^{2} - 4 x = 15

Move

15

to the left side

5 x^{2} - 4 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 ( - 15 )}{2 \cdot 5}

(- 4)^{2} - 4 \cdot 5 (- 15) = 279

x_{1, 2} = \frac{- ( - 4 ) \pm 2 79}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 79}{2 \cdot 5}, x_{2} = \frac{- ( - 4 ) - 2 79}{2 \cdot 5}

x = \frac{- ( - 4 ) + 2 79}{2 \cdot 5} : \frac{2 + 79}{5}

x = \frac{- ( - 4 ) - 2 79}{2 \cdot 5} : \frac{2 - 79}{5}

The solutions to the quadratic equation are:

x = \frac{2 + 79}{5}, x = \frac{2 - 79}{5}

y + y^{2} = 2

27 - x^{2} = 0

(x + 1)^{2} + 4 = 1

x, (x + y)^{2} + (x - y)^{2} = 0

x^{2} + 2 4^{2} = 3 0^{2}

What are the solutions to the equation x^2-4/5 x=3 ?
The solutions to the equation x^2-4/5 x=3 are x=(2+sqrt(79))/5 ,x=(2-sqrt(79))/5
Find the zeros of x^2-4/5 x=3
The zeros of x^2-4/5 x=3 are x=(2+sqrt(79))/5 ,x=(2-sqrt(79))/5