What are the solutions to the equation 32x^2-3x-14=(2x-1)^2 ?

The solutions to the equation 32x^2-3x-14=(2x-1)^2 are x= 5/7 ,x=-3/4

Find the zeros of 32x^2-3x-14=(2x-1)^2

The zeros of 32x^2-3x-14=(2x-1)^2 are x= 5/7 ,x=-3/4

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32 x^{2} - 3 x - 14 = (2 x - 1)^{2}

x = \frac{5}{7}, x = - \frac{3}{4}

Decimal

x = 0.71428 \dots, x = - 0.75

Solution steps

32 x^{2} - 3 x - 14 = (2 x - 1)^{2}

Expand

(2 x - 1)^{2} : 4 x^{2} - 4 x + 1

32 x^{2} - 3 x - 14 = 4 x^{2} - 4 x + 1

Move

1

to the left side

32 x^{2} - 3 x - 15 = 4 x^{2} - 4 x

Move

4 x

to the left side

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

Move

4 x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 28 ( - 15 )}{2 \cdot 28}

1^{2} - 4 \cdot 28 (- 15) = 41

x_{1, 2} = \frac{- 1 \pm 41}{2 \cdot 28}

Separate the solutions

x_{1} = \frac{- 1 + 41}{2 \cdot 28}, x_{2} = \frac{- 1 - 41}{2 \cdot 28}

x = \frac{- 1 + 41}{2 \cdot 28} : \frac{5}{7}

x = \frac{- 1 - 41}{2 \cdot 28} : - \frac{3}{4}

The solutions to the quadratic equation are:

x = \frac{5}{7}, x = - \frac{3}{4}

x, 4 x^{2} + 4 y^{2} - 16 x + 40 + 67 = 0

300 - 2 Q^{1} + Q^{2} = 60

6 x (4 x + 3) = 0

12 x^{2} - 320 x + 1500 = 0

4 x^{2} + 12 - 72 = 0

What are the solutions to the equation 32x^2-3x-14=(2x-1)^2 ?
The solutions to the equation 32x^2-3x-14=(2x-1)^2 are x= 5/7 ,x=-3/4
Find the zeros of 32x^2-3x-14=(2x-1)^2
The zeros of 32x^2-3x-14=(2x-1)^2 are x= 5/7 ,x=-3/4