What are the solutions to the equation (x+3)(x-1)=-3x+4 ?

The solutions to the equation (x+3)(x-1)=-3x+4 are x=(-5+sqrt(53))/2 ,x=(-5-sqrt(53))/2

Find the zeros of (x+3)(x-1)=-3x+4

The zeros of (x+3)(x-1)=-3x+4 are x=(-5+sqrt(53))/2 ,x=(-5-sqrt(53))/2

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(x + 3) (x - 1) = - 3 x + 4

x = \frac{- 5 + 53}{2}, x = \frac{- 5 - 53}{2}

Decimal Notation

x = 1.14005 \dots, x = - 6.14005 \dots

Solution steps

(x + 3) (x - 1) = - 3 x + 4

Expand

(x + 3) (x - 1) : x^{2} + 2 x - 3

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

Move

4

to the left side

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

Move

3 x

to the left side

x^{2} + 5 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 7) = 53

x_{1, 2} = \frac{- 5 \pm 53}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 53}{2 \cdot 1}, x_{2} = \frac{- 5 - 53}{2 \cdot 1}

x = \frac{- 5 + 53}{2 \cdot 1} : \frac{- 5 + 53}{2}

x = \frac{- 5 - 53}{2 \cdot 1} : \frac{- 5 - 53}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 53}{2}, x = \frac{- 5 - 53}{2}

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

x^{2} + x - 8

y, x = y^{2} - 3

(y + 6)^{2} - 72 = 0

x^{2} + 18 x + 85 = 0

What are the solutions to the equation (x+3)(x-1)=-3x+4 ?
The solutions to the equation (x+3)(x-1)=-3x+4 are x=(-5+sqrt(53))/2 ,x=(-5-sqrt(53))/2
Find the zeros of (x+3)(x-1)=-3x+4
The zeros of (x+3)(x-1)=-3x+4 are x=(-5+sqrt(53))/2 ,x=(-5-sqrt(53))/2