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

The solutions to the equation k(-3)=2x^2-5x+3 are x=(5+sqrt(-24k+1))/4 ,x=(5-sqrt(-24k+1))/4

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

The zeros of k(-3)=2x^2-5x+3 are x=(5+sqrt(-24k+1))/4 ,x=(5-sqrt(-24k+1))/4

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k (- 3) = 2 x^{2} - 5 x + 3

x = \frac{5 + - 24 k + 1}{4}, x = \frac{5 - - 24 k + 1}{4}

Solution steps

k (- 3) = 2 x^{2} - 5 x + 3

Expand

k (- 3) : - 3 k

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

Switch sides

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

Move

3 k

to the left side

2 x^{2} - 5 x + 3 + 3 k = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( 3 + 3 k )}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 (3 + 3 k) : - 24 k + 1

x_{1, 2} = \frac{- ( - 5 ) \pm - 24 k + 1}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + - 24 k + 1}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - - 24 k + 1}{2 \cdot 2}

x = \frac{- ( - 5 ) + - 24 k + 1}{2 \cdot 2} : \frac{5 + - 24 k + 1}{4}

x = \frac{- ( - 5 ) - - 24 k + 1}{2 \cdot 2} : \frac{5 - - 24 k + 1}{4}

The solutions to the quadratic equation are:

x = \frac{5 + - 24 k + 1}{4}, x = \frac{5 - - 24 k + 1}{4}

112 + 96 t - 16 t^{2} = 112

6 x^{2} + x = 36

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

x, 4 x^{2} + 3 x y - y^{2} = 0

x^{2} - 25 x + 58 = - 10 x + 8

What are the solutions to the equation k(-3)=2x^2-5x+3 ?
The solutions to the equation k(-3)=2x^2-5x+3 are x=(5+sqrt(-24k+1))/4 ,x=(5-sqrt(-24k+1))/4
Find the zeros of k(-3)=2x^2-5x+3
The zeros of k(-3)=2x^2-5x+3 are x=(5+sqrt(-24k+1))/4 ,x=(5-sqrt(-24k+1))/4