What are the solutions to the equation (2x-3)^2-5(2x-3)+6=0 ?

The solutions to the equation (2x-3)^2-5(2x-3)+6=0 are x=3,x= 5/2

Find the zeros of (2x-3)^2-5(2x-3)+6=0

The zeros of (2x-3)^2-5(2x-3)+6=0 are x=3,x= 5/2

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(2 x - 3)^{2} - 5 (2 x - 3) + 6 = 0

x = 3, x = \frac{5}{2}

Decimal Notation

x = 3, x = 2.5

Solution steps

(2 x - 3)^{2} - 5 (2 x - 3) + 6 = 0

Expand

(2 x - 3)^{2} - 5 (2 x - 3) + 6 : 4 x^{2} - 22 x + 30

4 x^{2} - 22 x + 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 4 \cdot 30}{2 \cdot 4}

(- 22)^{2} - 4 \cdot 4 \cdot 30 = 2

x_{1, 2} = \frac{- ( - 22 ) \pm 2}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 22 ) + 2}{2 \cdot 4}, x_{2} = \frac{- ( - 22 ) - 2}{2 \cdot 4}

x = \frac{- ( - 22 ) + 2}{2 \cdot 4} : 3

x = \frac{- ( - 22 ) - 2}{2 \cdot 4} : \frac{5}{2}

The solutions to the quadratic equation are:

x = 3, x = \frac{5}{2}

5 x^{2} - 4 x + 9 = 0

5 x^{2} + 20 x - 25 = 0

0 = 36 x^{2} - 24 x

6 x^{2} + 4 x = 0

z, (z + 3)^{3} + (z - m)^{3} = 2 (z^{3} + 6 m^{3})

What are the solutions to the equation (2x-3)^2-5(2x-3)+6=0 ?
The solutions to the equation (2x-3)^2-5(2x-3)+6=0 are x=3,x= 5/2
Find the zeros of (2x-3)^2-5(2x-3)+6=0
The zeros of (2x-3)^2-5(2x-3)+6=0 are x=3,x= 5/2