What are the solutions to the equation (2x-3)^2=11x-19 ?

The solutions to the equation (2x-3)^2=11x-19 are x=4,x= 7/4

Find the zeros of (2x-3)^2=11x-19

The zeros of (2x-3)^2=11x-19 are x=4,x= 7/4

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(2 x - 3)^{2} = 11 x - 19

x = 4, x = \frac{7}{4}

Solution steps

(2 x - 3)^{2} = 11 x - 19

Expand

(2 x - 3)^{2} : 4 x^{2} - 12 x + 9

4 x^{2} - 12 x + 9 = 11 x - 19

Move

19

to the left side

4 x^{2} - 12 x + 28 = 11 x

Move

11 x

to the left side

4 x^{2} - 23 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 23 ) \pm ( - 23 ) ^{2} - 4 \cdot 4 \cdot 28}{2 \cdot 4}

(- 23)^{2} - 4 \cdot 4 \cdot 28 = 9

x_{1, 2} = \frac{- ( - 23 ) \pm 9}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 23 ) + 9}{2 \cdot 4}, x_{2} = \frac{- ( - 23 ) - 9}{2 \cdot 4}

x = \frac{- ( - 23 ) + 9}{2 \cdot 4} : 4

x = \frac{- ( - 23 ) - 9}{2 \cdot 4} : \frac{7}{4}

The solutions to the quadratic equation are:

x = 4, x = \frac{7}{4}

949

- 7 + 2

- x^{2} + x + 1 = 1

\frac{2 - 8 i}{3 + 5 i}

9 w^{2} + 48 w + 64

What are the solutions to the equation (2x-3)^2=11x-19 ?
The solutions to the equation (2x-3)^2=11x-19 are x=4,x= 7/4
Find the zeros of (2x-3)^2=11x-19
The zeros of (2x-3)^2=11x-19 are x=4,x= 7/4