What are the solutions to the equation 2x(x-5)=3(x+2)-1 ?

The solutions to the equation 2x(x-5)=3(x+2)-1 are x=(13+sqrt(209))/4 ,x=(13-sqrt(209))/4

Find the zeros of 2x(x-5)=3(x+2)-1

The zeros of 2x(x-5)=3(x+2)-1 are x=(13+sqrt(209))/4 ,x=(13-sqrt(209))/4

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

x = \frac{13 + 209}{4}, x = \frac{13 - 209}{4}

Solution steps

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

Expand

2 x (x - 5) : 2 x^{2} - 10 x

Expand

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

2 x^{2} - 10 x = 3 x + 5

Move

5

to the left side

2 x^{2} - 10 x - 5 = 3 x

Move

3 x

to the left side

2 x^{2} - 13 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

(- 13)^{2} - 4 \cdot 2 (- 5) = 209

x_{1, 2} = \frac{- ( - 13 ) \pm 209}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 209}{2 \cdot 2}, x_{2} = \frac{- ( - 13 ) - 209}{2 \cdot 2}

x = \frac{- ( - 13 ) + 209}{2 \cdot 2} : \frac{13 + 209}{4}

x = \frac{- ( - 13 ) - 209}{2 \cdot 2} : \frac{13 - 209}{4}

The solutions to the quadratic equation are:

x = \frac{13 + 209}{4}, x = \frac{13 - 209}{4}

∣ 7 x - 7 ∣ = 0

109 < - 3 + 8 b

r + 0.2 < - 0.7

16 (4)

12 a^{2} + 5 a - 2

What are the solutions to the equation 2x(x-5)=3(x+2)-1 ?
The solutions to the equation 2x(x-5)=3(x+2)-1 are x=(13+sqrt(209))/4 ,x=(13-sqrt(209))/4
Find the zeros of 2x(x-5)=3(x+2)-1
The zeros of 2x(x-5)=3(x+2)-1 are x=(13+sqrt(209))/4 ,x=(13-sqrt(209))/4