What are the solutions to the equation (x-4)(2x-3)=12 ?

The solutions to the equation (x-4)(2x-3)=12 are x= 11/2 ,x=0

Find the zeros of (x-4)(2x-3)=12

The zeros of (x-4)(2x-3)=12 are x= 11/2 ,x=0

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(x - 4) (2 x - 3) = 12

x = \frac{11}{2}, x = 0

Decimal Notation

x = 5.5, x = 0

Solution steps

(x - 4) (2 x - 3) = 12

Expand

(x - 4) (2 x - 3) : 2 x^{2} - 11 x + 12

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

Move

12

to the left side

2 x^{2} - 11 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 \cdot 0 = 11

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

Separate the solutions

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

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

x = \frac{- ( - 11 ) - 11}{2 \cdot 2} : 0

The solutions to the quadratic equation are:

x = \frac{11}{2}, x = 0

4 x^{2} = 24 x - 35

x, f = x^{2} + 2 x y + y^{2}

y, y^{2} - 2 x y^{2} = x^{3} y + 5 x^{2} y^{2} - y

(x + 2)^{2} = 12 (x + 2)

0 = - x^{2} - 4 x + 12

What are the solutions to the equation (x-4)(2x-3)=12 ?
The solutions to the equation (x-4)(2x-3)=12 are x= 11/2 ,x=0
Find the zeros of (x-4)(2x-3)=12
The zeros of (x-4)(2x-3)=12 are x= 11/2 ,x=0