What are the solutions to the equation (2x-3)(x+4)=6 ?

The solutions to the equation (2x-3)(x+4)=6 are x=2,x=-9/2

Find the zeros of (2x-3)(x+4)=6

The zeros of (2x-3)(x+4)=6 are x=2,x=-9/2

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(2 x - 3) (x + 4) = 6

x = 2, x = - \frac{9}{2}

Decimal Notation

x = 2, x = - 4.5

Solution steps

(2 x - 3) (x + 4) = 6

Expand

(2 x - 3) (x + 4) : 2 x^{2} + 5 x - 12

2 x^{2} + 5 x - 12 = 6

Move

6

to the left side

2 x^{2} + 5 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 18 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 18) = 13

x_{1, 2} = \frac{- 5 \pm 13}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 13}{2 \cdot 2}, x_{2} = \frac{- 5 - 13}{2 \cdot 2}

x = \frac{- 5 + 13}{2 \cdot 2} : 2

x = \frac{- 5 - 13}{2 \cdot 2} : - \frac{9}{2}

The solutions to the quadratic equation are:

x = 2, x = - \frac{9}{2}

x^{2} - 10 x = - 9 x - 1

(x + 6) x = 720

x^{2} + 18 x = 15

d = 0.006 (1 - \frac{( ( 1.82 ) - ( d ) ( 1500 ) )}{- 3})^{2}

0 = a (x - 3)^{2} + 1

What are the solutions to the equation (2x-3)(x+4)=6 ?
The solutions to the equation (2x-3)(x+4)=6 are x=2,x=-9/2
Find the zeros of (2x-3)(x+4)=6
The zeros of (2x-3)(x+4)=6 are x=2,x=-9/2