What are the solutions to the equation 4x^2-4x+7x-7=2(x-1) ?

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

Find the zeros of 4x^2-4x+7x-7=2(x-1)

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

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4 x^{2} - 4 x + 7 x - 7 = 2 (x - 1)

x = 1, x = - \frac{5}{4}

Decimal Notation

x = 1, x = - 1.25

Solution steps

4 x^{2} - 4 x + 7 x - 7 = 2 (x - 1)

Expand

2 (x - 1) : 2 x - 2

4 x^{2} + 3 x - 7 = 2 x - 2

Move

2

to the left side

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

Move

2 x

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 4 (- 5) = 9

x_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 1 + 9}{2 \cdot 4}, x_{2} = \frac{- 1 - 9}{2 \cdot 4}

x = \frac{- 1 + 9}{2 \cdot 4} : 1

x = \frac{- 1 - 9}{2 \cdot 4} : - \frac{5}{4}

The solutions to the quadratic equation are:

x = 1, x = - \frac{5}{4}

2^{4 x} \leq 20

f = \frac{( 3 ) ^{3} + 11}{( 3 ) ^{3}}

3 (3 - x) \geq x + 1

\frac{12 n ^{2} - 29 n - 8}{28 n ^{2} - 5 n - 3}

∣ 4 - 4 x ∣ \leq 1

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