What are the solutions to the equation (x+5)(2x-3)=9 ?

The solutions to the equation (x+5)(2x-3)=9 are x=(-7+sqrt(241))/4 ,x=(-7-sqrt(241))/4

Find the zeros of (x+5)(2x-3)=9

The zeros of (x+5)(2x-3)=9 are x=(-7+sqrt(241))/4 ,x=(-7-sqrt(241))/4

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(x + 5) (2 x - 3) = 9

x = \frac{- 7 + 241}{4}, x = \frac{- 7 - 241}{4}

Decimal Notation

x = 2.13104 \dots, x = - 5.63104 \dots

Solution steps

(x + 5) (2 x - 3) = 9

Expand

(x + 5) (2 x - 3) : 2 x^{2} + 7 x - 15

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

Move

9

to the left side

2 x^{2} + 7 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 24 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 24) = 241

x_{1, 2} = \frac{- 7 \pm 241}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 241}{2 \cdot 2}, x_{2} = \frac{- 7 - 241}{2 \cdot 2}

x = \frac{- 7 + 241}{2 \cdot 2} : \frac{- 7 + 241}{4}

x = \frac{- 7 - 241}{2 \cdot 2} : \frac{- 7 - 241}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 241}{4}, x = \frac{- 7 - 241}{4}

2 x^{2} + 7 x - 18 = x^{2}

2 0^{2} + 2 \cdot 32.2 = v^{2} + 50 (5 - 2)^{2}

x, 3.1396 (1 - x^{2}) = (π) (x)^{2}

- t^{2} + 27 t + 3 = - 2 t^{2} + 25 t + 27

- 6 x^{2} = 4

What are the solutions to the equation (x+5)(2x-3)=9 ?
The solutions to the equation (x+5)(2x-3)=9 are x=(-7+sqrt(241))/4 ,x=(-7-sqrt(241))/4
Find the zeros of (x+5)(2x-3)=9
The zeros of (x+5)(2x-3)=9 are x=(-7+sqrt(241))/4 ,x=(-7-sqrt(241))/4