What are the solutions to the equation (x-4)(x-1)=28 ?

The solutions to the equation (x-4)(x-1)=28 are x=8,x=-3

Find the zeros of (x-4)(x-1)=28

The zeros of (x-4)(x-1)=28 are x=8,x=-3

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(x - 4) (x - 1) = 28

x = 8, x = - 3

Solution steps

(x - 4) (x - 1) = 28

Expand

(x - 4) (x - 1) : x^{2} - 5 x + 4

x^{2} - 5 x + 4 = 28

Move

28

to the left side

x^{2} - 5 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 24) = 11

x_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 1}

x = \frac{- ( - 5 ) + 11}{2 \cdot 1} : 8

x = \frac{- ( - 5 ) - 11}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 8, x = - 3

x^{2} - 3.5 x - 15 = 0

3 g^{2} - 12 g = - 4

d^{2} = 45

66 x^{2} - 55 x = 0

5 x^{2} + 8 = 0

What are the solutions to the equation (x-4)(x-1)=28 ?
The solutions to the equation (x-4)(x-1)=28 are x=8,x=-3
Find the zeros of (x-4)(x-1)=28
The zeros of (x-4)(x-1)=28 are x=8,x=-3