What are the solutions to the equation 2x^2+3x=4x+12 ?

The solutions to the equation 2x^2+3x=4x+12 are x=(1+sqrt(97))/4 ,x=(1-sqrt(97))/4

Find the zeros of 2x^2+3x=4x+12

The zeros of 2x^2+3x=4x+12 are x=(1+sqrt(97))/4 ,x=(1-sqrt(97))/4

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2 x^{2} + 3 x = 4 x + 12

x = \frac{1 + 97}{4}, x = \frac{1 - 97}{4}

Solution steps

2 x^{2} + 3 x = 4 x + 12

Move

12

to the left side

2 x^{2} + 3 x - 12 = 4 x

Move

4 x

to the left side

2 x^{2} - x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 12 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 12) = 97

x_{1, 2} = \frac{- ( - 1 ) \pm 97}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 97}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 97}{2 \cdot 2}

x = \frac{- ( - 1 ) + 97}{2 \cdot 2} : \frac{1 + 97}{4}

x = \frac{- ( - 1 ) - 97}{2 \cdot 2} : \frac{1 - 97}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 97}{4}, x = \frac{1 - 97}{4}

2 x^{3} + x y^{2} - 2 x^{2} y - y^{3}

\frac{3}{10} x - \frac{1}{5} = - 2

p + 9 < 7

\frac{( x - 2 ) ( x - 3 )}{6} - \frac{( x - 1 ) ^{2}}{4} = 2 - x

\frac{7}{10} \cdot \frac{6}{9}

What are the solutions to the equation 2x^2+3x=4x+12 ?
The solutions to the equation 2x^2+3x=4x+12 are x=(1+sqrt(97))/4 ,x=(1-sqrt(97))/4
Find the zeros of 2x^2+3x=4x+12
The zeros of 2x^2+3x=4x+12 are x=(1+sqrt(97))/4 ,x=(1-sqrt(97))/4