What are the solutions to the equation (2x+1)(x-4)-10=0 ?

The solutions to the equation (2x+1)(x-4)-10=0 are x=(7+sqrt(161))/4 ,x=(7-sqrt(161))/4

Find the zeros of (2x+1)(x-4)-10=0

The zeros of (2x+1)(x-4)-10=0 are x=(7+sqrt(161))/4 ,x=(7-sqrt(161))/4

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(2 x + 1) (x - 4) - 10 = 0

x = \frac{7 + 161}{4}, x = \frac{7 - 161}{4}

Decimal Notation

x = 4.92214 \dots, x = - 1.42214 \dots

Solution steps

(2 x + 1) (x - 4) - 10 = 0

Expand

(2 x + 1) (x - 4) - 10 : 2 x^{2} - 7 x - 14

2 x^{2} - 7 x - 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 14 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 14) = 161

x_{1, 2} = \frac{- ( - 7 ) \pm 161}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 161}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 161}{2 \cdot 2}

x = \frac{- ( - 7 ) + 161}{2 \cdot 2} : \frac{7 + 161}{4}

x = \frac{- ( - 7 ) - 161}{2 \cdot 2} : \frac{7 - 161}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 161}{4}, x = \frac{7 - 161}{4}

x, x^{2} + 7 x = 0

0.544 x^{2} - 4.9 x + 7.385 = 0

4 x^{2} + 4 x + 6 = 0

3 x^{2} - 18 x + 24 = 0

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

What are the solutions to the equation (2x+1)(x-4)-10=0 ?
The solutions to the equation (2x+1)(x-4)-10=0 are x=(7+sqrt(161))/4 ,x=(7-sqrt(161))/4
Find the zeros of (2x+1)(x-4)-10=0
The zeros of (2x+1)(x-4)-10=0 are x=(7+sqrt(161))/4 ,x=(7-sqrt(161))/4