What are the solutions to the equation 4x^2-5x=x^2+28 ?

The solutions to the equation 4x^2-5x=x^2+28 are x=4,x=-7/3

Find the zeros of 4x^2-5x=x^2+28

The zeros of 4x^2-5x=x^2+28 are x=4,x=-7/3

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4 x^{2} - 5 x = x^{2} + 28

x = 4, x = - \frac{7}{3}

Decimal Notation

x = 4, x = - 2.33333 \dots

Solution steps

4 x^{2} - 5 x = x^{2} + 28

Move

28

to the left side

4 x^{2} - 5 x - 28 = x^{2}

Move

x^{2}

to the left side

3 x^{2} - 5 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 28 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 28) = 19

x_{1, 2} = \frac{- ( - 5 ) \pm 19}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 19}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 19}{2 \cdot 3}

x = \frac{- ( - 5 ) + 19}{2 \cdot 3} : 4

x = \frac{- ( - 5 ) - 19}{2 \cdot 3} : - \frac{7}{3}

The solutions to the quadratic equation are:

x = 4, x = - \frac{7}{3}

3 m^{2} + 13 = m^{2} + 397

2 x^{2} y - 5 x^{2} - 2 y + 5

\frac{3}{4} = \frac{n}{12} n

\frac{1}{2} f - 4 = 1

- \frac{23}{2} (\frac{1}{100 0 ^{2}}) + \frac{23}{2}

What are the solutions to the equation 4x^2-5x=x^2+28 ?
The solutions to the equation 4x^2-5x=x^2+28 are x=4,x=-7/3
Find the zeros of 4x^2-5x=x^2+28
The zeros of 4x^2-5x=x^2+28 are x=4,x=-7/3