What are the solutions to the equation (2x-1)(x+3)=4 ?

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

Find the zeros of (2x-1)(x+3)=4

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

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(2 x - 1) (x + 3) = 4

x = 1, x = - \frac{7}{2}

Solution steps

(2 x - 1) (x + 3) = 4

Expand

(2 x - 1) (x + 3) : 2 x^{2} + 5 x - 3

2 x^{2} + 5 x - 3 = 4

Move

4

to the left side

2 x^{2} + 5 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 7) = 9

x_{1, 2} = \frac{- 5 \pm 9}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 9}{2 \cdot 2}, x_{2} = \frac{- 5 - 9}{2 \cdot 2}

x = \frac{- 5 + 9}{2 \cdot 2} : 1

x = \frac{- 5 - 9}{2 \cdot 2} : - \frac{7}{2}

The solutions to the quadratic equation are:

x = 1, x = - \frac{7}{2}

x^{2} - \frac{2}{3} x = \frac{3}{5}

3 x^{2} + 7 = 4

x^{2} = a, a > 0

y, x = 3 lo g_{10} (y^{2})

x, M (x) = 4 x^{2} - 12 x + 8

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