What are the solutions to the equation (2x-3/4)(x-1/2)=31 ?

The solutions to the equation (2x-3/4)(x-1/2)=31 are x= 35/8 ,x=-7/2

Find the zeros of (2x-3/4)(x-1/2)=31

The zeros of (2x-3/4)(x-1/2)=31 are x= 35/8 ,x=-7/2

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(2 x - \frac{3}{4}) (x - \frac{1}{2}) = 31

x = \frac{35}{8}, x = - \frac{7}{2}

Solution steps

(2 x - \frac{3}{4}) (x - \frac{1}{2}) = 31

Expand

(2 x - \frac{3}{4}) (x - \frac{1}{2}) : 2 x^{2} - \frac{7}{4} x + \frac{3}{8}

2 x^{2} - \frac{7}{4} x + \frac{3}{8} = 31

Move

31

to the left side

- \frac{245}{8} - \frac{7}{4} x + 2 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} - \frac{7 x}{4} - \frac{245}{8} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - \frac{7}{4} ) \pm ( - \frac{7}{4} ) ^{2} - 4 \cdot 2 ( - \frac{245}{8} )}{2 \cdot 2}

(- \frac{7}{4})^{2} - 4 \cdot 2 (- \frac{245}{8}) = \frac{63}{4}

x_{1, 2} = \frac{- ( - \frac{7}{4} ) \pm \frac{63}{4}}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - \frac{7}{4} ) + \frac{63}{4}}{2 \cdot 2}, x_{2} = \frac{- ( - \frac{7}{4} ) - \frac{63}{4}}{2 \cdot 2}

x = \frac{- ( - \frac{7}{4} ) + \frac{63}{4}}{2 \cdot 2} : \frac{35}{8}

x = \frac{- ( - \frac{7}{4} ) - \frac{63}{4}}{2 \cdot 2} : - \frac{7}{2}

The solutions to the quadratic equation are:

x = \frac{35}{8}, x = - \frac{7}{2}

\frac{142.17 ( 1.2 )}{( 32.11 ) ( 79.1 )}

9 \leq 2 m + 2 - 3

a^{2} + 6 a + 9 = 0

15 \cdot \frac{1}{2}

lo g_{25} (\frac{1}{5}) = x

What are the solutions to the equation (2x-3/4)(x-1/2)=31 ?
The solutions to the equation (2x-3/4)(x-1/2)=31 are x= 35/8 ,x=-7/2
Find the zeros of (2x-3/4)(x-1/2)=31
The zeros of (2x-3/4)(x-1/2)=31 are x= 35/8 ,x=-7/2