What are the solutions to the equation (2x+3)^2-4(2x+3)-12=0 ?

The solutions to the equation (2x+3)^2-4(2x+3)-12=0 are x= 3/2 ,x=-5/2

Find the zeros of (2x+3)^2-4(2x+3)-12=0

The zeros of (2x+3)^2-4(2x+3)-12=0 are x= 3/2 ,x=-5/2

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(2 x + 3)^{2} - 4 (2 x + 3) - 12 = 0

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

Decimal Notation

x = 1.5, x = - 2.5

Solution steps

(2 x + 3)^{2} - 4 (2 x + 3) - 12 = 0

Expand

(2 x + 3)^{2} - 4 (2 x + 3) - 12 : 4 x^{2} + 4 x - 15

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 4 (- 15) = 16

x_{1, 2} = \frac{- 4 \pm 16}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 4 + 16}{2 \cdot 4}, x_{2} = \frac{- 4 - 16}{2 \cdot 4}

x = \frac{- 4 + 16}{2 \cdot 4} : \frac{3}{2}

x = \frac{- 4 - 16}{2 \cdot 4} : - \frac{5}{2}

The solutions to the quadratic equation are:

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

3 x + 1 > 34 - 4 x

30 + 4 x = 12 x

x^{3} + 2 x^{2} - 24 x

\frac{3}{4} - \frac{1}{8}

∣ x ∣ = 8

What are the solutions to the equation (2x+3)^2-4(2x+3)-12=0 ?
The solutions to the equation (2x+3)^2-4(2x+3)-12=0 are x= 3/2 ,x=-5/2
Find the zeros of (2x+3)^2-4(2x+3)-12=0
The zeros of (2x+3)^2-4(2x+3)-12=0 are x= 3/2 ,x=-5/2