What are the solutions to the equation 15x^2-4=17x ?

The solutions to the equation 15x^2-4=17x are x= 4/3 ,x=-1/5

Find the zeros of 15x^2-4=17x

The zeros of 15x^2-4=17x are x= 4/3 ,x=-1/5

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15 x^{2} - 4 = 17 x

x = \frac{4}{3}, x = - \frac{1}{5}

Decimal Notation

x = 1.33333 \dots, x = - 0.2

Solution steps

15 x^{2} - 4 = 17 x

Move

17 x

to the left side

15 x^{2} - 4 - 17 x = 0

Write in the standard form

a x^{2} + b x + c = 0

15 x^{2} - 17 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 15 ( - 4 )}{2 \cdot 15}

(- 17)^{2} - 4 \cdot 15 (- 4) = 23

x_{1, 2} = \frac{- ( - 17 ) \pm 23}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- ( - 17 ) + 23}{2 \cdot 15}, x_{2} = \frac{- ( - 17 ) - 23}{2 \cdot 15}

x = \frac{- ( - 17 ) + 23}{2 \cdot 15} : \frac{4}{3}

x = \frac{- ( - 17 ) - 23}{2 \cdot 15} : - \frac{1}{5}

The solutions to the quadratic equation are:

x = \frac{4}{3}, x = - \frac{1}{5}

2 - \frac{1}{2} n = 3 n + 16

3.643 \cdot 1 0^{- 1}

(- 5)^{- 2}

(\frac{2 x ^{3}}{yz})^{3}

\frac{1}{17}

What are the solutions to the equation 15x^2-4=17x ?
The solutions to the equation 15x^2-4=17x are x= 4/3 ,x=-1/5
Find the zeros of 15x^2-4=17x
The zeros of 15x^2-4=17x are x= 4/3 ,x=-1/5