What are the solutions to the equation (2x+3)(x+4)=228 ?

The solutions to the equation (2x+3)(x+4)=228 are x=8,x=-27/2

Find the zeros of (2x+3)(x+4)=228

The zeros of (2x+3)(x+4)=228 are x=8,x=-27/2

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(2 x + 3) (x + 4) = 228

x = 8, x = - \frac{27}{2}

Solution steps

(2 x + 3) (x + 4) = 228

Expand

(2 x + 3) (x + 4) : 2 x^{2} + 11 x + 12

2 x^{2} + 11 x + 12 = 228

Move

228

to the left side

2 x^{2} + 11 x - 216 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 ( - 216 )}{2 \cdot 2}

1 1^{2} - 4 \cdot 2 (- 216) = 43

x_{1, 2} = \frac{- 11 \pm 43}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 11 + 43}{2 \cdot 2}, x_{2} = \frac{- 11 - 43}{2 \cdot 2}

x = \frac{- 11 + 43}{2 \cdot 2} : 8

x = \frac{- 11 - 43}{2 \cdot 2} : - \frac{27}{2}

The solutions to the quadratic equation are:

x = 8, x = - \frac{27}{2}

3 (4 x^{2} - 1) + 6 (- 2 x + 3) = - 7

5 x^{2} - 2 x - 35 = 0

4 n^{2} - 20 n + 16 = 0

220 - (16.5 - 2 x)^{2} = 99.9

5 x^{2} + 20 x

What are the solutions to the equation (2x+3)(x+4)=228 ?
The solutions to the equation (2x+3)(x+4)=228 are x=8,x=-27/2
Find the zeros of (2x+3)(x+4)=228
The zeros of (2x+3)(x+4)=228 are x=8,x=-27/2