What are the solutions to the equation 2x^2-3x=2x+15 ?

The solutions to the equation 2x^2-3x=2x+15 are x=(5+sqrt(145))/4 ,x=(5-sqrt(145))/4

Find the zeros of 2x^2-3x=2x+15

The zeros of 2x^2-3x=2x+15 are x=(5+sqrt(145))/4 ,x=(5-sqrt(145))/4

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2 x^{2} - 3 x = 2 x + 15

x = \frac{5 + 145}{4}, x = \frac{5 - 145}{4}

Solution steps

2 x^{2} - 3 x = 2 x + 15

Move

15

to the left side

2 x^{2} - 3 x - 15 = 2 x

Move

2 x

to the left side

2 x^{2} - 5 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 15) = 145

x_{1, 2} = \frac{- ( - 5 ) \pm 145}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 145}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 145}{2 \cdot 2}

x = \frac{- ( - 5 ) + 145}{2 \cdot 2} : \frac{5 + 145}{4}

x = \frac{- ( - 5 ) - 145}{2 \cdot 2} : \frac{5 - 145}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 145}{4}, x = \frac{5 - 145}{4}

n (x) = - 2 x^{2}

2 x + x^{2} = - 13

6 x^{2} - 16 x - 6 = 0

10 x^{2} + 25 x = 15

- 3 x^{2} - 6 x - 5 = 0

What are the solutions to the equation 2x^2-3x=2x+15 ?
The solutions to the equation 2x^2-3x=2x+15 are x=(5+sqrt(145))/4 ,x=(5-sqrt(145))/4
Find the zeros of 2x^2-3x=2x+15
The zeros of 2x^2-3x=2x+15 are x=(5+sqrt(145))/4 ,x=(5-sqrt(145))/4