What are the solutions to the equation (2x-7)(4x+3)=112 ?

The solutions to the equation (2x-7)(4x+3)=112 are x=(11+sqrt(1185))/8 ,x=(11-sqrt(1185))/8

Find the zeros of (2x-7)(4x+3)=112

The zeros of (2x-7)(4x+3)=112 are x=(11+sqrt(1185))/8 ,x=(11-sqrt(1185))/8

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(2 x - 7) (4 x + 3) = 112

x = \frac{11 + 1185}{8}, x = \frac{11 - 1185}{8}

Decimal Notation

x = 5.67797 \dots, x = - 2.92797 \dots

Solution steps

(2 x - 7) (4 x + 3) = 112

Expand

(2 x - 7) (4 x + 3) : 8 x^{2} - 22 x - 21

8 x^{2} - 22 x - 21 = 112

Move

112

to the left side

8 x^{2} - 22 x - 133 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 8 ( - 133 )}{2 \cdot 8}

(- 22)^{2} - 4 \cdot 8 (- 133) = 21185

x_{1, 2} = \frac{- ( - 22 ) \pm 2 1185}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 22 ) + 2 1185}{2 \cdot 8}, x_{2} = \frac{- ( - 22 ) - 2 1185}{2 \cdot 8}

x = \frac{- ( - 22 ) + 2 1185}{2 \cdot 8} : \frac{11 + 1185}{8}

x = \frac{- ( - 22 ) - 2 1185}{2 \cdot 8} : \frac{11 - 1185}{8}

The solutions to the quadratic equation are:

x = \frac{11 + 1185}{8}, x = \frac{11 - 1185}{8}

u^{2} - 15 u - 34 = 0

a, a^{2} b - c = d

4 k^{2} - 64 = 0

4 - 81 x^{2} = 0

m^{2} + 2 m - 120 = 0

What are the solutions to the equation (2x-7)(4x+3)=112 ?
The solutions to the equation (2x-7)(4x+3)=112 are x=(11+sqrt(1185))/8 ,x=(11-sqrt(1185))/8
Find the zeros of (2x-7)(4x+3)=112
The zeros of (2x-7)(4x+3)=112 are x=(11+sqrt(1185))/8 ,x=(11-sqrt(1185))/8