What are the solutions to the equation (2x+3)^2-4x=13-7x ?

The solutions to the equation (2x+3)^2-4x=13-7x are x= 1/4 ,x=-4

Find the zeros of (2x+3)^2-4x=13-7x

The zeros of (2x+3)^2-4x=13-7x are x= 1/4 ,x=-4

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(2 x + 3)^{2} - 4 x = 13 - 7 x

x = \frac{1}{4}, x = - 4

Decimal Notation

x = 0.25, x = - 4

Solution steps

(2 x + 3)^{2} - 4 x = 13 - 7 x

Expand

(2 x + 3)^{2} - 4 x : 4 x^{2} + 8 x + 9

4 x^{2} + 8 x + 9 = 13 - 7 x

Move

7 x

to the left side

4 x^{2} + 15 x + 9 = 13

Move

13

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

1 5^{2} - 4 \cdot 4 (- 4) = 17

x_{1, 2} = \frac{- 15 \pm 17}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 15 + 17}{2 \cdot 4}, x_{2} = \frac{- 15 - 17}{2 \cdot 4}

x = \frac{- 15 + 17}{2 \cdot 4} : \frac{1}{4}

x = \frac{- 15 - 17}{2 \cdot 4} : - 4

The solutions to the quadratic equation are:

x = \frac{1}{4}, x = - 4

- 4 x^{2} + 32 x + 80 = 0

x^{2} + 8 x = - 12 + 0

- 2 x^{2} + 5 x + 8 = 0

5 x^{2} + 29 x + 20 = 0

y = x^{2} - 9 x + 20, x

What are the solutions to the equation (2x+3)^2-4x=13-7x ?
The solutions to the equation (2x+3)^2-4x=13-7x are x= 1/4 ,x=-4
Find the zeros of (2x+3)^2-4x=13-7x
The zeros of (2x+3)^2-4x=13-7x are x= 1/4 ,x=-4