What are the solutions to the equation x(3x-5)=8 ?

The solutions to the equation x(3x-5)=8 are x= 8/3 ,x=-1

Find the zeros of x(3x-5)=8

The zeros of x(3x-5)=8 are x= 8/3 ,x=-1

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x (3 x - 5) = 8

x = \frac{8}{3}, x = - 1

Decimal Notation

x = 2.66666 \dots, x = - 1

Solution steps

x (3 x - 5) = 8

Expand

x (3 x - 5) : 3 x^{2} - 5 x

3 x^{2} - 5 x = 8

Move

8

to the left side

3 x^{2} - 5 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 8 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 8) = 11

x_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 3}

x = \frac{- ( - 5 ) + 11}{2 \cdot 3} : \frac{8}{3}

x = \frac{- ( - 5 ) - 11}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

x = \frac{8}{3}, x = - 1

\frac{10}{11} - \frac{1}{4}

a^{4} - 2 a^{2} b^{2} + b^{4}

\frac{x}{14} \geq 1 - \frac{x}{28}

(\frac{1}{16})^{3 x - 4} < 6 4^{x - 1}

(8 r)^{\frac{1}{3}} (2 r^{\frac{1}{2}})

What are the solutions to the equation x(3x-5)=8 ?
The solutions to the equation x(3x-5)=8 are x= 8/3 ,x=-1
Find the zeros of x(3x-5)=8
The zeros of x(3x-5)=8 are x= 8/3 ,x=-1