What are the solutions to the equation (x+1)(x-3)=2 ?

The solutions to the equation (x+1)(x-3)=2 are x=1+sqrt(6),x=1-sqrt(6)

Find the zeros of (x+1)(x-3)=2

The zeros of (x+1)(x-3)=2 are x=1+sqrt(6),x=1-sqrt(6)

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

x = 1 + 6, x = 1 - 6

Decimal Notation

x = 3.44948 \dots, x = - 1.44948 \dots

Solution steps

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

Expand

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

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

Move

2

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 5) = 26

x_{1, 2} = \frac{- ( - 2 ) \pm 2 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 6}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 6}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 6}{2 \cdot 1} : 1 + 6

x = \frac{- ( - 2 ) - 2 6}{2 \cdot 1} : 1 - 6

The solutions to the quadratic equation are:

x = 1 + 6, x = 1 - 6

4 = 2 x^{2} + 3

4.6 y^{2} + 3.5 y - 1.75 = 0

\frac{2}{3} (x - 4) (x + 5) = 1

y, x^{2} y + x^{3} y^{2} = 2

t, d (t) = - 9.81 t^{2} + 10 t m

What are the solutions to the equation (x+1)(x-3)=2 ?
The solutions to the equation (x+1)(x-3)=2 are x=1+sqrt(6),x=1-sqrt(6)
Find the zeros of (x+1)(x-3)=2
The zeros of (x+1)(x-3)=2 are x=1+sqrt(6),x=1-sqrt(6)