What are the solutions to the equation x(x+4)+3(x-1)=5(x-1)+3 ?

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

Find the zeros of x(x+4)+3(x-1)=5(x-1)+3

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

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x (x + 4) + 3 (x - 1) = 5 (x - 1) + 3

x = - 1 + 2, x = - 1 - 2

Decimal Notation

x = 0.41421 \dots, x = - 2.41421 \dots

Solution steps

x (x + 4) + 3 (x - 1) = 5 (x - 1) + 3

Expand

x (x + 4) + 3 (x - 1) : x^{2} + 7 x - 3

Expand

5 (x - 1) + 3 : 5 x - 2

x^{2} + 7 x - 3 = 5 x - 2

Move

2

to the left side

x^{2} + 7 x - 1 = 5 x

Move

5 x

to the left side

x^{2} + 2 x - 1 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 + 2, x = - 1 - 2

\frac{x ^{2}}{6} - \frac{x}{2} = 3 (x - 1)

s, 11 s^{2} - 26 s - 61 = 0

t, tP = - t^{2} + 17 t + 190

2 + 3 x - 4 x^{2} = 0

x^{2} + 7 x - 30 = 0

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