What are the solutions to the equation 15x-10=6x(x-2)+(-x+3) ?

The solutions to the equation 15x-10=6x(x-2)+(-x+3) are x=(14+sqrt(118))/6 ,x=(14-sqrt(118))/6

Find the zeros of 15x-10=6x(x-2)+(-x+3)

The zeros of 15x-10=6x(x-2)+(-x+3) are x=(14+sqrt(118))/6 ,x=(14-sqrt(118))/6

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15 x - 10 = 6 x (x - 2) + (- x + 3)

x = \frac{14 + 118}{6}, x = \frac{14 - 118}{6}

Decimal Notation

x = 4.14379 \dots, x = 0.52286 \dots

Solution steps

15 x - 10 = 6 x (x - 2) + (- x + 3)

Expand

6 x (x - 2) + (- x + 3) : 6 x^{2} - 13 x + 3

15 x - 10 = 6 x^{2} - 13 x + 3

Switch sides

6 x^{2} - 13 x + 3 = 15 x - 10

Move

10

to the left side

6 x^{2} - 13 x + 13 = 15 x

Move

15 x

to the left side

6 x^{2} - 28 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 6 \cdot 13}{2 \cdot 6}

(- 28)^{2} - 4 \cdot 6 \cdot 13 = 2118

x_{1, 2} = \frac{- ( - 28 ) \pm 2 118}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 28 ) + 2 118}{2 \cdot 6}, x_{2} = \frac{- ( - 28 ) - 2 118}{2 \cdot 6}

x = \frac{- ( - 28 ) + 2 118}{2 \cdot 6} : \frac{14 + 118}{6}

x = \frac{- ( - 28 ) - 2 118}{2 \cdot 6} : \frac{14 - 118}{6}

The solutions to the quadratic equation are:

x = \frac{14 + 118}{6}, x = \frac{14 - 118}{6}

2 y^{2} + 5 y = 0

x, f^{- 1} = x^{2} + 7 x + 10

4 x^{2} + 24 x + 45 = 0

x^{2} + x = a^{2} + a

28 x^{2} + 3 x - 18 = 0

What are the solutions to the equation 15x-10=6x(x-2)+(-x+3) ?
The solutions to the equation 15x-10=6x(x-2)+(-x+3) are x=(14+sqrt(118))/6 ,x=(14-sqrt(118))/6
Find the zeros of 15x-10=6x(x-2)+(-x+3)
The zeros of 15x-10=6x(x-2)+(-x+3) are x=(14+sqrt(118))/6 ,x=(14-sqrt(118))/6