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

The solutions to the equation (2x-3)(x-2)=5 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4

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

The zeros of (2x-3)(x-2)=5 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4

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(2 x - 3) (x - 2) = 5

x = \frac{7 + 41}{4}, x = \frac{7 - 41}{4}

Decimal Notation

x = 3.35078 \dots, x = 0.14921 \dots

Solution steps

(2 x - 3) (x - 2) = 5

Expand

(2 x - 3) (x - 2) : 2 x^{2} - 7 x + 6

2 x^{2} - 7 x + 6 = 5

Move

5

to the left side

2 x^{2} - 7 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 1 = 41

x_{1, 2} = \frac{- ( - 7 ) \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 41}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 41}{2 \cdot 2}

x = \frac{- ( - 7 ) + 41}{2 \cdot 2} : \frac{7 + 41}{4}

x = \frac{- ( - 7 ) - 41}{2 \cdot 2} : \frac{7 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 41}{4}, x = \frac{7 - 41}{4}

100 - x^{2} = x^{2} + 4 x + 4

a^{2} + 1 6^{2} = 1 8^{2}

n^{2} + 3 n - 28 = 0

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

6 x^{2} + 9 x + 2 = - 2 - 4 x - 3 x^{2}

What are the solutions to the equation (2x-3)(x-2)=5 ?
The solutions to the equation (2x-3)(x-2)=5 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4
Find the zeros of (2x-3)(x-2)=5
The zeros of (2x-3)(x-2)=5 are x=(7+sqrt(41))/4 ,x=(7-sqrt(41))/4