What are the solutions to the equation 5x^2-18x+4=12x+39 ?

The solutions to the equation 5x^2-18x+4=12x+39 are x=7,x=-1

Find the zeros of 5x^2-18x+4=12x+39

The zeros of 5x^2-18x+4=12x+39 are x=7,x=-1

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5 x^{2} - 18 x + 4 = 12 x + 39

x = 7, x = - 1

Solution steps

5 x^{2} - 18 x + 4 = 12 x + 39

Move

39

to the left side

5 x^{2} - 18 x - 35 = 12 x

Move

12 x

to the left side

5 x^{2} - 30 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 5 ( - 35 )}{2 \cdot 5}

(- 30)^{2} - 4 \cdot 5 (- 35) = 40

x_{1, 2} = \frac{- ( - 30 ) \pm 40}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 30 ) + 40}{2 \cdot 5}, x_{2} = \frac{- ( - 30 ) - 40}{2 \cdot 5}

x = \frac{- ( - 30 ) + 40}{2 \cdot 5} : 7

x = \frac{- ( - 30 ) - 40}{2 \cdot 5} : - 1

The solutions to the quadratic equation are:

x = 7, x = - 1

x^{2} = 43

(\frac{125}{64})^{- \frac{2}{3}}

x^{3} + x^{2} + 3 x + 3

36 r^{3} + 30 r^{2} - 30 r - 25

2 x < - 3 x + 15

What are the solutions to the equation 5x^2-18x+4=12x+39 ?
The solutions to the equation 5x^2-18x+4=12x+39 are x=7,x=-1
Find the zeros of 5x^2-18x+4=12x+39
The zeros of 5x^2-18x+4=12x+39 are x=7,x=-1