What are the solutions to the equation 2(4a+4)^{(2)}+4(4a+4)-6=0 ?

The solutions to the equation 2(4a+4)^{(2)}+4(4a+4)-6=0 are a=-3/4 ,a=-7/4

Find the zeros of 2(4a+4)^{(2)}+4(4a+4)-6=0

The zeros of 2(4a+4)^{(2)}+4(4a+4)-6=0 are a=-3/4 ,a=-7/4

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2 (4 a + 4)^{(2)} + 4 (4 a + 4) - 6 = 0

a = - \frac{3}{4}, a = - \frac{7}{4}

Decimal Notation

a = - 0.75, a = - 1.75

Solution steps

2 (4 a + 4)^{(2)} + 4 (4 a + 4) - 6 = 0

Expand

2 (4 a + 4)^{2} + 4 (4 a + 4) - 6 : 32 a^{2} + 80 a + 42

32 a^{2} + 80 a + 42 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 32 \cdot 42}{2 \cdot 32}

8 0^{2} - 4 \cdot 32 \cdot 42 = 32

a_{1, 2} = \frac{- 80 \pm 32}{2 \cdot 32}

Separate the solutions

a_{1} = \frac{- 80 + 32}{2 \cdot 32}, a_{2} = \frac{- 80 - 32}{2 \cdot 32}

a = \frac{- 80 + 32}{2 \cdot 32} : - \frac{3}{4}

a = \frac{- 80 - 32}{2 \cdot 32} : - \frac{7}{4}

The solutions to the quadratic equation are:

a = - \frac{3}{4}, a = - \frac{7}{4}

x^{2} + 18 x - 63

5^{12}

∣ x ∣ < 6

5 c^{2} + 30 c + 45

\frac{v ^{2} - 1}{v + 1} \leq 0

What are the solutions to the equation 2(4a+4)^{(2)}+4(4a+4)-6=0 ?
The solutions to the equation 2(4a+4)^{(2)}+4(4a+4)-6=0 are a=-3/4 ,a=-7/4
Find the zeros of 2(4a+4)^{(2)}+4(4a+4)-6=0
The zeros of 2(4a+4)^{(2)}+4(4a+4)-6=0 are a=-3/4 ,a=-7/4