What are the solutions to the equation 4x^2-7x+21=9x+6 ?

The solutions to the equation 4x^2-7x+21=9x+6 are x= 5/2 ,x= 3/2

Find the zeros of 4x^2-7x+21=9x+6

The zeros of 4x^2-7x+21=9x+6 are x= 5/2 ,x= 3/2

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4 x^{2} - 7 x + 21 = 9 x + 6

x = \frac{5}{2}, x = \frac{3}{2}

Decimal Notation

x = 2.5, x = 1.5

Solution steps

4 x^{2} - 7 x + 21 = 9 x + 6

Move

6

to the left side

4 x^{2} - 7 x + 15 = 9 x

Move

9 x

to the left side

4 x^{2} - 16 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 15}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 \cdot 15 = 4

x_{1, 2} = \frac{- ( - 16 ) \pm 4}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4}{2 \cdot 4}, x_{2} = \frac{- ( - 16 ) - 4}{2 \cdot 4}

x = \frac{- ( - 16 ) + 4}{2 \cdot 4} : \frac{5}{2}

x = \frac{- ( - 16 ) - 4}{2 \cdot 4} : \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = \frac{3}{2}

- 0.1 x^{2} + 0.8 x + 8 = 0

- 4 x^{3} y (- 15 x y^{5})

\frac{5}{3} + 2

4 \leq \frac{2}{3} x + 2 < 10

\frac{5}{6} x + \frac{7}{9} - \frac{2}{3} x = \frac{2}{9} x - \frac{5}{9}

What are the solutions to the equation 4x^2-7x+21=9x+6 ?
The solutions to the equation 4x^2-7x+21=9x+6 are x= 5/2 ,x= 3/2
Find the zeros of 4x^2-7x+21=9x+6
The zeros of 4x^2-7x+21=9x+6 are x= 5/2 ,x= 3/2