What are the solutions to the equation ((x^2))/6-((3x))/6 =3(x-5) ?

The solutions to the equation ((x^2))/6-((3x))/6 =3(x-5) are x=15,x=6

Find the zeros of ((x^2))/6-((3x))/6 =3(x-5)

The zeros of ((x^2))/6-((3x))/6 =3(x-5) are x=15,x=6

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\frac{( x ^{2} )}{6} - \frac{( 3 x )}{6} = 3 (x - 5)

x = 15, x = 6

Solution steps

\frac{( x ^{2} )}{6} - \frac{( 3 x )}{6} = 3 (x - 5)

Multiply both sides by

6

x^{2} - 3 x = 18 (x - 5)

Expand

18 (x - 5) : 18 x - 90

x^{2} - 3 x = 18 x - 90

Move

90

to the left side

x^{2} - 3 x + 90 = 18 x

Move

18 x

to the left side

x^{2} - 21 x + 90 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 1 \cdot 90}{2 \cdot 1}

(- 21)^{2} - 4 \cdot 1 \cdot 90 = 9

x_{1, 2} = \frac{- ( - 21 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 21 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 21 ) - 9}{2 \cdot 1}

x = \frac{- ( - 21 ) + 9}{2 \cdot 1} : 15

x = \frac{- ( - 21 ) - 9}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

x = 15, x = 6

4 x^{2} - 4 a^{2} \cdot x + (a^{4} - b^{4}) = 0

x, x^{2} - 10 x + y^{2} - 20 y = - 125

x, z = (5 x + 3) (6 x + 2 y)

\frac{3 \cdot x ^{2} - 2 x + 1}{2} = 0

x^{2} + 1.2 x = 0

What are the solutions to the equation ((x^2))/6-((3x))/6 =3(x-5) ?
The solutions to the equation ((x^2))/6-((3x))/6 =3(x-5) are x=15,x=6
Find the zeros of ((x^2))/6-((3x))/6 =3(x-5)
The zeros of ((x^2))/6-((3x))/6 =3(x-5) are x=15,x=6