What are the solutions to the equation (2x-5)(x+1)=5 ?

The solutions to the equation (2x-5)(x+1)=5 are x=(3+sqrt(89))/4 ,x=(3-sqrt(89))/4

Find the zeros of (2x-5)(x+1)=5

The zeros of (2x-5)(x+1)=5 are x=(3+sqrt(89))/4 ,x=(3-sqrt(89))/4

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(2 x - 5) (x + 1) = 5

x = \frac{3 + 89}{4}, x = \frac{3 - 89}{4}

Decimal Notation

x = 3.10849 \dots, x = - 1.60849 \dots

Solution steps

(2 x - 5) (x + 1) = 5

Expand

(2 x - 5) (x + 1) : 2 x^{2} - 3 x - 5

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

Move

5

to the left side

2 x^{2} - 3 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 10 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 10) = 89

x_{1, 2} = \frac{- ( - 3 ) \pm 89}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 89}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 89}{2 \cdot 2}

x = \frac{- ( - 3 ) + 89}{2 \cdot 2} : \frac{3 + 89}{4}

x = \frac{- ( - 3 ) - 89}{2 \cdot 2} : \frac{3 - 89}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 89}{4}, x = \frac{3 - 89}{4}

\frac{3}{1000} = \frac{1}{2} (0.005) (x - 3 - 0.5)^{2}

6 x^{2} - 20 x + 6 = 0

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

x^{2} + 6 x + 9 = \frac{49}{4}

\frac{1}{2} x^{2} - 12 = - 37

What are the solutions to the equation (2x-5)(x+1)=5 ?
The solutions to the equation (2x-5)(x+1)=5 are x=(3+sqrt(89))/4 ,x=(3-sqrt(89))/4
Find the zeros of (2x-5)(x+1)=5
The zeros of (2x-5)(x+1)=5 are x=(3+sqrt(89))/4 ,x=(3-sqrt(89))/4