What are the solutions to the equation 4q^2+8q-1=4 ?

The solutions to the equation 4q^2+8q-1=4 are q= 1/2 ,q=-5/2

Find the zeros of 4q^2+8q-1=4

The zeros of 4q^2+8q-1=4 are q= 1/2 ,q=-5/2

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4 q^{2} + 8 q - 1 = 4

q = \frac{1}{2}, q = - \frac{5}{2}

Decimal Notation

q = 0.5, q = - 2.5

Solution steps

4 q^{2} + 8 q - 1 = 4

Move

4

to the left side

4 q^{2} + 8 q - 5 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

8^{2} - 4 \cdot 4 (- 5) = 12

q_{1, 2} = \frac{- 8 \pm 12}{2 \cdot 4}

Separate the solutions

q_{1} = \frac{- 8 + 12}{2 \cdot 4}, q_{2} = \frac{- 8 - 12}{2 \cdot 4}

q = \frac{- 8 + 12}{2 \cdot 4} : \frac{1}{2}

q = \frac{- 8 - 12}{2 \cdot 4} : - \frac{5}{2}

The solutions to the quadratic equation are:

q = \frac{1}{2}, q = - \frac{5}{2}

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

28 + 19 v = - 8 + 15 v

45 s^{4} - 36 s^{3} - 10 s + 8

x^{3} - 5 x^{2} - 14 x > 0

6 x^{2} - 2

What are the solutions to the equation 4q^2+8q-1=4 ?
The solutions to the equation 4q^2+8q-1=4 are q= 1/2 ,q=-5/2
Find the zeros of 4q^2+8q-1=4
The zeros of 4q^2+8q-1=4 are q= 1/2 ,q=-5/2