What are the solutions to the equation x^2+5x-33=4x+9 ?

The solutions to the equation x^2+5x-33=4x+9 are x=6,x=-7

Find the zeros of x^2+5x-33=4x+9

The zeros of x^2+5x-33=4x+9 are x=6,x=-7

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x^{2} + 5 x - 33 = 4 x + 9

x = 6, x = - 7

Solution steps

x^{2} + 5 x - 33 = 4 x + 9

Move

9

to the left side

x^{2} + 5 x - 42 = 4 x

Move

4 x

to the left side

x^{2} + x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 \cdot 1}, x_{2} = \frac{- 1 - 13}{2 \cdot 1}

x = \frac{- 1 + 13}{2 \cdot 1} : 6

x = \frac{- 1 - 13}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 6, x = - 7

5 x - 9 = 15 + 11 x

163 > \frac{3}{4} x^{2} - 33 x > 45

(x + 4)^{2} + 2 x = 0

4 (5) - 6 (- 3) \geq - 18

- 2 + 8 - 4 z \geq 8

What are the solutions to the equation x^2+5x-33=4x+9 ?
The solutions to the equation x^2+5x-33=4x+9 are x=6,x=-7
Find the zeros of x^2+5x-33=4x+9
The zeros of x^2+5x-33=4x+9 are x=6,x=-7